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Compact Operators and Riesz Schauder Theory

1 · Prerequisites

2 · Summary

The page begins from the published definition of a bounded linear operator and the metric notion of compactness, and defines a compact operator by the compact closure of the image of every bounded set, with the closed unit ball sufficient. The sequential characterisation under Dependent Choice, the compactness of bounded finite-rank operators, the two-sided ideal property, closure under linear combinations and norm closure for Banach targets follow in that order; the latter yields the notion of an approximable operator, for which only the approximation-property direction into a fixed Banach target is recorded, and deliberately not the converse. Schauder's compact-adjoint theorem is proved in both directions, and a compact operator is shown to send weakly convergent sequences to norm convergent ones.

The second half develops Riesz--Schauder theory for $I - K$ with $K$ compact on a Banach space: finite-dimensional kernel, closed range with its distance estimate, the stabilisation of the kernel and range chains and the resulting direct-sum decomposition with an invertible restriction of $I - K$ , and then the Fredholm alternative with its adjoint solvability condition and the equality of the finite defect dimensions. Neumann series and small perturbations of bounded inverses, the spectrum and resolvent vocabulary, and the Riesz--Schauder spectral theorem for compact operators follow: every nonzero spectral value is an eigenvalue of finite algebraic multiplicity, only finitely many spectral values lie outside any positive radius, and the spectrum of an infinite-dimensional operator contains zero; the countability corollary records that the only possible accumulation point is zero.

The page closes with the Fredholm index: the definition through a finite dimension kernel, closed range and finite-dimensional cokernel, the splitting into finite-dimensional defects with a parametrix, the compact-remainder estimate forcing closed range, Atkinson's parametrix characterisation, additivity of the index through an exact six-term sequence, local constancy through a finite-dimensional Schur complement, invariance under compact perturbations along the connected path $t \mapsto T + t K$ , and the index-zero corollary for $λ I - K$ with $λ \neq 0$ .

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Compact linear operator

Definition

Let $X$ and $Y$ be normed spaces over the same scalar field $K$ , read in the real case from A norm on a real vector space, the induced metric, and the dictionary with the metric axioms and in the complex case from Real and complex scalar conventions for normed spaces. A linear map $T : X \to Y$ (Linear map between vector spaces over the same field) is a compact operator when the image of every bounded subset of $X$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space) has compact closure in $Y$ (Open cover, subcover, compact metric space, and compact subset of a metric space); explicitly, for every bounded $A \subseteq X$ the closure $\overline{T (A)}$ in $Y$ is a compact subset of $Y$ . The set of compact operators $X \to Y$ is written $K (X, Y)$ .

The closed unit ball suffices. Put ${\overline{B}}_{X} : = {x \in X : ∥ x ∥ \leq 1}$ (Open ball, closed ball and sphere in a metric space). Then $T$ is compact if and only if $\overline{T ({\overline{B}}_{X})}$ is a compact subset of $Y$ .

Indeed, if $T$ is compact then ${\overline{B}}_{X}$ is bounded, because ${\overline{B}}_{X} \subseteq B (0, 2)$ while $B (0, 2)$ is bounded and a subset of a bounded set is bounded, so $T ({\overline{B}}_{X})$ has compact closure. Conversely assume $\overline{T ({\overline{B}}_{X})}$ compact and let $A \subseteq X$ be bounded. If $A = \emptyset$ then $T (A) = \emptyset$ , whose closure is empty and hence compact. Otherwise $A \subseteq B (x_{0}, r)$ for some $x_{0} \in X$ and real $r > 0$ , so every $x \in A$ satisfies $∥ x ∥ \leq ∥ x_{0} ∥ + r = : R$ ; if $R > 0$ then $A \subseteq R {\overline{B}}_{X}$ and if $R = 0$ then $A \subseteq {0}$ , so in either case $T (A) \subseteq R \overline{T ({\overline{B}}_{X})}$ . Scalar multiplication by $R$ is continuous (Vector addition and scalar multiplication are continuous in a normed space), so $R \overline{T ({\overline{B}}_{X})}$ is a compact subset of $Y$ (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism), hence closed (A compact subset of a metric space is closed and bounded); therefore $\overline{T (A)} \subseteq R \overline{T ({\overline{B}}_{X})}$ is a closed subset of a compact set, hence compact, and $T$ is compact.

A compact operator is bounded. If $T$ is compact then the compact set $\overline{T ({\overline{B}}_{X})}$ is bounded (A compact subset of a metric space is closed and bounded), so there is a real $C \geq 0$ with $∥ T x ∥ \leq C$ for every $x \in {\overline{B}}_{X}$ and hence $∥ T x ∥ \leq C ∥ x ∥$ for every $x \in X$ ; thus $T$ is a bounded linear operator (A bounded linear operator between normed spaces). This is a consequence of compactness, not a hypothesis of the definition.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Dependent choice implies countable choice

Statement

In ZF, the Axiom of Dependent Choice implies the Axiom of Countable Choice: every at most countable family of nonempty sets has a choice function (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Choice function, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Facts & Assumptions

[A1]

$D C$ : for every nonempty set $X$ , every relation $R \subseteq X \times X$ entire on $X$ and every $a \in X$ there is a function $x : N \to X$ with $x_{0} = a$ and $x_{n} R x_{n + 1}$ for every $n$ (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The natural numbers $N$ (von Neumann), A function is a relation $f$ with $(a, b) \in f$ and $(a, c) \in f$ implying $b = c$ ; $f : A \to B$ , the value $f (a)$ , domain and codomain).

[A2]

${A C}_{ω}$ : for every family $(E_{n})_{n \in N}$ of nonempty sets there is a function $f$ with domain $N$ and $f (n) \in E_{n}$ for all $n$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). A choice function on the set $E : = {E_{n} : n \in N}$ instead has domain $E$ and selects an element of each of its members (Choice function).

[A3]

For sets $A$ and $B$ the collection of functions $A \to B$ is a set (For sets $A$ and $B$ the collection of all functions $A \to B$ is a set, being a subset of $P (A \times B)$ , The power set $P (x) = {z : z \subseteq x}$ , The Axiom Schema of Separation: for each formula $φ$ , $\forall \overset{ˉ}{p} \forall x \exists y \forall z (z \in y \leftrightarrow (z \in x \land φ (z, \overset{ˉ}{p})))$ , The Kuratowski ordered pair $(a, b) : = {{a}, {a, b}}$ ); unions and intersections of indexed families are available ( $⋃_{i \in I} A_{i} : = ⋃ {A_{i} : i \in I}$ , and $⋂_{i \in I} A_{i} : = ⋂ {A_{i} : i \in I}$ for $I \neq \emptyset$ ), induction on $N$ is available (The principle of mathematical induction, The natural numbers $N$ (von Neumann)), and every nonempty subset of $N$ has a least element (The well-ordering principle); a natural number is the set of its predecessors, so $⋃_{n \in N} n = N$ .

Proof

technique · direct

Given: $D C$ and a family $(E_{n})_{n \in N}$ of nonempty sets.

1.1

Let $H$ be the set of all functions $s$ with $dom s \in N$ and $s (j) \in E_{j}$ for every $j < dom s$ : this is a set by [A3] applied inside $⋃_{n \in N} E_{n}$ , and the empty function lies in $H$ , so $H \neq \emptyset$ .

2.1

Define $R \subseteq H \times H$ by $s R t$ if and only if $dom t = dom s + 1$ and $t (j) = s (j)$ for every $j < dom s$ . Then $R$ is entire on $H$ : given $s \in H$ , the set $E_{dom s}$ is nonempty, and for any $x \in E_{dom s}$ the function $t : = s \cup {(dom s, x)}$ lies in $H$ with $s R t$ .

step 1.1A1

3.1

By [A1] applied to $H$ , $R$ and the empty function there is a sequence $(s_{n})_{n \in N}$ in $H$ with $s_{0} = \emptyset$ and $s_{n} R s_{n + 1}$ for every $n$ .

step 1.1step 2.1A1

4.1

For every $n$ one has $dom s_{n} = n$ , by induction on $n$ from [A3]: $dom s_{0} = dom \emptyset = 0$ , and $dom s_{n + 1} = dom s_{n} + 1 = n + 1$ .

step 3.1A3

4.2

The union $f : = ⋃_{n \in N} s_{n}$ is a function: if $(j, y)$ and $(j, y^{'})$ lie in $f$ , they lie in $s_{m}$ and $s_{m^{'}}$ for some $m, m^{'}$ , and with $m \leq m^{'}$ the relation gives $s_{m} \subseteq s_{m^{'}}$ , so $y = y^{'}$ .

step 3.1A3

5.1

Its domain is $N$ : $dom f = ⋃_{n} dom s_{n} = ⋃_{n} n = N$ by [step 4.1] and [A3].

step 4.1A3

5.2

For every $n \in N$ one has $f (n) = s_{n + 1} (n) \in E_{n}$ : the point $n$ lies in $dom s_{n + 1}$ by [step 4.1] and $s_{n + 1} \in H$ with $dom s_{n + 1} = n + 1 > n$ .

step 4.1

6.1

Put $E : = {E_{n} : n \in N}$ . For each $E \in E$ , the set ${n : E_{n} = E}$ is a nonempty subset of $N$ , so let $n (E)$ be its least element and define $c (E) : = f (n (E))$ . Then $c$ has domain $E$ and $c (E) \in E_{n (E)} = E$ , so $c$ is a choice function on the set of members even when the indexed family has repetitions.

step 5.2A2A3

7.1

Hence $f$ is a function with domain $N$ and $f (n) \in E_{n}$ for every $n$ , which is exactly the indexed conclusion of [A2], while $c$ is the corresponding choice function on the set of member sets. Since the family was arbitrary, $D C$ implies ${A C}_{ω}$ .

step 4.2step 5.1step 5.2step 6.1A2∎

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Sequential characterization of compact operators

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Let $X$ and $Y$ be normed spaces over the same scalar field (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, Real and complex scalar conventions for normed spaces) and let $T : X \to Y$ be a bounded linear operator (A bounded linear operator between normed spaces). Then $T$ is compact (Compact linear operator) if and only if every bounded sequence $(x_{k})$ in $X$ , that is every function $x : N \to X$ with bounded range (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), has a subsequence $(x_{k_{j}})$ , the index map being strictly increasing (Countably compact, sequentially compact and limit point compact metric spaces, A strictly increasing index map satisfies $n_{k} \geq k$ ), for which $(T x_{k_{j}})$ converges in the norm metric of $Y$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

Facts & Assumptions

[A1]

$T$ is compact exactly when $\overline{T ({\overline{B}}_{X})}$ is a compact subset of $Y$ , where ${\overline{B}}_{X} = {x \in X : ∥ x ∥ \leq 1}$ (Compact linear operator).

[A2]

Assume ${A C}_{ω}$ and $D C$ . For a metric space $(M, d)$ , compactness, countable compactness, limit point compactness, sequential compactness and "complete and totally bounded" are equivalent; only "sequentially compact implies totally bounded" spends DC and only "complete and totally bounded implies compact" spends ${A C}_{ω}$ , so every other implication is a theorem of ZF (For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice).

[A3]

In ZF, $D C$ implies ${A C}_{ω}$ (Dependent choice implies countable choice); and $D C$ is the statement that for every nonempty set $X$ , every relation $R$ on $X$ entire on $X$ and every $a \in X$ there is $x : N \to X$ with $x_{0} = a$ and $x_{n} R x_{n + 1}$ for all $n$ (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[A4]

A subset $A$ of a metric space is bounded when $A = \emptyset$ or $A \subseteq B (x_{0}, r)$ for some point $x_{0}$ and real $r > 0$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open ball, closed ball and sphere in a metric space).

[A5]

In a normed space, $∥ x ∥ \leq R$ implies $x \in R {\overline{B}}_{X}$ ; convergence is metric convergence for $d (y, y^{'}) = ∥ y - y^{'} ∥$ ; a set $F \subseteq Y$ is closed exactly when $F = \overline{F}$ ; for nonempty $A$ the closure is $\overline{A} = {y : d (y, A) = 0}$ ; and the closure of $A$ is contained in every closed set containing $A$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

Proof

technique · direct

Given: $D C$ , normed spaces $X, Y$ over one scalar field, a bounded linear $T : X \to Y$ , and the notation ${\overline{B}}_{X} = {x : ∥ x ∥ \leq 1}$ .

1.1

Suppose $T$ compact and let $(x_{k})$ be a bounded sequence in $X$ . By [A4] the range ${x_{k}}$ is empty or contained in some ball $B (x_{0}, r)$ ; because $∥ x_{k} ∥ \leq ∥ x_{k} - x_{0} ∥ + ∥ x_{0} ∥$ for every $k$ , there is a real $R \geq 0$ with $∥ x_{k} ∥ \leq R$ for all $k$ (if the range is empty take $R = 0$ ), and then $T x_{k} \in R \overline{T ({\overline{B}}_{X})}$ for every $k$ , because $x_{k} \in R {\overline{B}}_{X}$ and $T$ is linear.

A4A5algebra

1.2

Conversely assume every bounded sequence in $X$ has a subsequence whose $T$ -images converge, and put $C : = \overline{T ({\overline{B}}_{X})}$ . Let $(y_{k})$ be a sequence in $C$ . For each $k$ the set $S_{k} : = {x \in {\overline{B}}_{X} : ∥ y_{k} - T x ∥ < 1 / (k + 1)}$ is nonempty, because $y_{k} \in \overline{T ({\overline{B}}_{X})}$ , the set $T ({\overline{B}}_{X})$ is nonempty, and [A5] then gives a point of $T ({\overline{B}}_{X})$ within $1 / (k + 1)$ of $y_{k}$ ; selecting $x_{k} \in S_{k}$ for every $k$ is a countable selection from nonempty sets, which [A3] licenses.

A3A5

2.1

By [A1] the set $\overline{T ({\overline{B}}_{X})}$ is compact, and multiplication by the scalar $R$ is continuous, so $R \overline{T ({\overline{B}}_{X})}$ is a compact subset of $Y$ (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism); by [A2] its metric subspace is sequentially compact, so the sequence $(T x_{k})$ of [step 1.1] has a subsequence converging in $Y$ .

step 1.1A1A2

2.2

The sequence $(x_{k})$ of [step 1.2] is bounded, since its range lies in ${\overline{B}}_{X}$ ; by hypothesis some subsequence $(x_{k_{j}})$ has $T x_{k_{j}} \to y$ for some $y \in Y$ , and since every $T x_{k_{j}}$ lies in $T ({\overline{B}}_{X}) \subseteq C$ while $C$ is closed, [A5] gives $y \in C$ .

step 1.2A4A5

3.1

Therefore compactness of $T$ implies the stated sequential property for every bounded sequence.

step 2.1

3.2

Along that subsequence $∥ y_{k_{j}} - y ∥ \leq ∥ y_{k_{j}} - T x_{k_{j}} ∥ + ∥ T x_{k_{j}} - y ∥ < 1 / (k_{j} + 1) + ∥ T x_{k_{j}} - y ∥$ , and both terms tend to $0$ , so $y_{k_{j}} \to y$ with $y \in C$ .

step 1.2step 2.2algebra

4.1

Thus every sequence in $C$ has a subsequence converging in $C$ , that is, $C$ is sequentially compact; by [A2] $C$ is compact, and then $T$ is compact by [A1].

step 3.2A1A2∎

LemmaStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Bounded finite rank operators are compact

Statement

Let $X$ and $Y$ be normed spaces over the same scalar field and let $T : X \to Y$ be a bounded linear operator (A bounded linear operator between normed spaces) whose range $T (X)$ admits an ordered basis of finite length (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis). Then $T$ is compact (Compact linear operator).

Facts & Assumptions

[A1]

$T$ is compact exactly when $\overline{T ({\overline{B}}_{X})}$ is compact, where ${\overline{B}}_{X} = {x \in X : ∥ x ∥ \leq 1}$ (Compact linear operator); the operator norm satisfies $∥ T x ∥ \leq ∥ T ∥ ∥ x ∥$ for every $x$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces).

[A2]

A subspace $W$ of a normed space $V$ admitting an ordered basis of finite length is a closed subset of $V$ (A finite-dimensional normed subspace is closed, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis), and its closed unit ball ${w \in W : ∥ w ∥ \leq 1}$ is compact (The closed unit ball is compact if and only if the normed space is finite-dimensional).

[A3]

A continuous image of a compact set is compact (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism), and scalar multiplication is continuous on a normed space (Vector addition and scalar multiplication are continuous in a normed space).

[A4]

A compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded); a closed subset of a compact topological space is a compact subset (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact).

Proof

technique · direct

Given: Normed spaces $X, Y$ over one scalar field and a bounded linear $T : X \to Y$ whose range $R : = T (X)$ admits an ordered basis of finite length.

1.1

Every point of $T ({\overline{B}}_{X})$ lies in $R$ , and $∥ T x ∥ \leq ∥ T ∥$ for every $x \in {\overline{B}}_{X}$ by [A1]; writing $s : = ∥ T ∥$ , this says $T ({\overline{B}}_{X}) \subseteq s {\overline{B}}_{R}$ , where ${\overline{B}}_{R} = {v \in R : ∥ v ∥ \leq 1}$ .

A1algebra

1.2

By [A2] the set ${\overline{B}}_{R}$ is compact and $R$ is closed in $Y$ .

2.1

The set $s {\overline{B}}_{R}$ is the image of the compact set ${\overline{B}}_{R}$ under the continuous map $v \mapsto s v$ , so it is a compact subset of $Y$ by [A3]; it is therefore closed in $Y$ by [A4].

step 1.2A3A4

3.1

Since $T ({\overline{B}}_{X}) \subseteq s {\overline{B}}_{R}$ by [step 1.1], the closure $\overline{T ({\overline{B}}_{X})}$ is contained in the closed set $s {\overline{B}}_{R}$ ; being a closed subset of the compact space $s {\overline{B}}_{R}$ , it is compact by [A4].

step 1.1step 2.1A4

4.1

By [A1] compactness of $\overline{T ({\overline{B}}_{X})}$ is exactly compactness of $T$ , so $T$ is compact.

step 3.1A1∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Compositions with a compact operator are compact

Statement

Let $W$ , $X$ , $Y$ and $Z$ be normed spaces over the same scalar field. If $T : X \to Y$ is compact (Compact linear operator) and $A : W \to X$ and $B : Y \to Z$ are bounded linear operators (A bounded linear operator between normed spaces), then the composites $T A : W \to Y$ and $B T : X \to Z$ are compact.

Facts & Assumptions

[A1]

A bounded linear operator is continuous and satisfies $∥ S w ∥ \leq ∥ S ∥ ∥ w ∥$ for all $w$ (For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent, The operator norm as the least bound and as the unit-sphere or unit-ball supremum); $T$ is compact exactly when $\overline{T (E)}$ is compact for every bounded $E \subseteq X$ , in particular for $E = {\overline{B}}_{X} = {x : ∥ x ∥ \leq 1}$ (Compact linear operator).

[A2]

A subset $E$ of a metric space is bounded when $E = \emptyset$ or $E \subseteq B (x_{0}, r)$ for some point $x_{0}$ and real $r > 0$ ; a subset of a bounded set is bounded (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[A3]

A continuous image of a compact subset is compact (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism); scalar multiplication by a fixed scalar is continuous (Vector addition and scalar multiplication are continuous in a normed space); a compact subset of a metric space is closed, and a closed subset of a compact metric space is compact (A compact subset of a metric space is closed and bounded, A closed subset of a compact metric space is compact).

Proof

technique · direct

Given: Normed spaces $W, X, Y, Z$ over one scalar field, a compact $T : X \to Y$ , and bounded linear $A : W \to X$ , $B : Y \to Z$ .

1.1

If $E \subseteq W$ is bounded and nonempty, say $E \subseteq B (w_{0}, r)$ , then $∥ A w ∥ \leq ∥ A ∥ (∥ w_{0} ∥ + r)$ for every $w \in E$ by [A1], so $A (E) \subseteq B (0, ∥ A ∥ (∥ w_{0} ∥ + r) + 1)$ is bounded; and $A (\emptyset) = \emptyset$ is bounded, so $A$ carries bounded sets to bounded sets.

A1A2algebra

1.2

If $E = \emptyset$ , take $R = 1$ , so $E \subseteq R {\overline{B}}_{X}$ holds immediately. If $E$ is a bounded subset of $X$ with $E \neq \emptyset$ and $E \subseteq B (x_{0}, R_{0})$ , then $∥ x ∥ \leq ∥ x_{0} ∥ + R_{0} = : R$ for every $x \in E$ by [A2] and the triangle inequality, so $E \subseteq R {\overline{B}}_{X}$ ; the same holds in any normed space.

A2algebra

1.3

The set $B (\overline{T ({\overline{B}}_{X})})$ is compact: $\overline{T ({\overline{B}}_{X})}$ is compact by [A1], and $B$ is continuous by [A1], so the image under $B$ is compact by [A3].

A1A3

2.1

For every bounded $E \subseteq W$ the image $A (E)$ is bounded by [step 1.1], so $\overline{T (A (E))}$ is compact by [A1]; hence $T A$ is compact.

step 1.1A1

2.2

For every bounded $E \subseteq X$ , [step 1.2] gives $E \subseteq R {\overline{B}}_{X}$ for some real $R \geq 0$ , so $T (E) \subseteq R \overline{T ({\overline{B}}_{X})}$ and hence $B (T (E)) \subseteq R B (\overline{T ({\overline{B}}_{X})})$ , which is compact by [step 1.3] and [A3] and therefore closed; thus $\overline{B (T (E))} \subseteq R B (\overline{T ({\overline{B}}_{X})})$ is a closed subset of a compact set, hence compact, and $B T$ is compact.

step 1.2step 1.3A1A3

3.1

Both composites $T A$ and $B T$ are therefore compact.

step 2.1step 2.2∎

LemmaStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Linear combinations of compact operators are compact

Statement

Let $X$ and $Y$ be normed spaces over the same scalar field. Then the compact operators $X \to Y$ (Compact linear operator) form a linear subspace of $B (X, Y)$ (A bounded linear operator between normed spaces): the zero operator is compact, and if $S, T : X \to Y$ are compact and $λ$ is a scalar, then $S + T$ and $λ S$ are compact.

Consequently, if $K : X \to X$ is a compact endomorphism, if $m \geq 1$ is a natural number and $a_{1}, \dots, a_{m}$ are scalars, then the polynomial $\sum_{k = 1}^{m} a_{k} K^{k} = a_{1} K + a_{2} K^{2} + \dots + a_{m} K^{m}$ is compact.

Facts & Assumptions

[A1]

$S$ is compact exactly when $\overline{S (E)}$ is compact for every bounded $E$ ; in particular $\overline{S ({\overline{B}}_{X})}$ is compact for the closed unit ball (Compact linear operator), and a bounded linear operator such as $K$ satisfies $∥ K w ∥ \leq ∥ K ∥ ∥ w ∥$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A2]

The zero operator has range ${0}$ , and the space ${0}$ admits the empty ordered basis of finite length, so the zero operator is compact (Bounded finite rank operators are compact, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[A3]

If $T$ is compact and $C$ is bounded linear, then $T C$ and $C T$ are compact (Compositions with a compact operator are compact).

[A4]

Addition $+ : Y \times Y \to Y$ and scalar multiplication are continuous (Vector addition and scalar multiplication are continuous in a normed space); a finite product of compact spaces is compact (A product of finitely many compact spaces is compact in the product topology); a continuous image of a compact set is compact (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism); a compact subset of a metric space is closed, and a closed subset of a compact metric space is compact (A compact subset of a metric space is closed and bounded, A closed subset of a compact metric space is compact).

Proof

technique · direct

Given: Normed spaces $X, Y$ over one scalar field, compact operators $S, T : X \to Y$ , a scalar $λ$ , and a compact endomorphism $K : X \to X$ .

1.1

The zero operator is compact by [A2].

1.2

For bounded $E \subseteq X$ the sets $\overline{S (E)}$ and $\overline{T (E)}$ are compact by [A1], so their product is compact in $Y \times Y$ by [A4] and its image under the continuous addition map is compact by [A4]; since $(S + T) (E) \subseteq \overline{S (E)} + \overline{T (E)}$ , its closure is a closed subset of that compact image, hence compact, and $S + T$ is compact.

A1A4algebra

1.3

For bounded $E \subseteq X$ the set $λ \overline{S (E)}$ is the image of the compact set $\overline{S (E)}$ under the continuous map $y \mapsto λ y$ , hence compact by [A4]; since $(λ S) (E) \subseteq λ \overline{S (E)}$ , its closure is compact by [A4], so $λ S$ is compact.

A1A4

1.4

For every natural $j \geq 1$ the power $K^{j}$ is compact: $K^{1} = K$ is compact, and if $K^{j}$ is compact then $K^{j + 1} = K \circ K^{j}$ is compact by [A3].

2.1

By [step 1.1], [step 1.2] and [step 1.3] the compact operators $X \to Y$ contain the zero operator and are closed under addition and scalar multiplication, so they form a linear subspace of $B (X, Y)$ .

step 1.1step 1.2step 1.3

3.1

Let $K$ be compact and $a_{1}, \dots, a_{m}$ scalars with $m \geq 1$ . Each $K^{k}$ with $1 \leq k \leq m$ is compact by [step 1.4]; by induction on $m$ using [step 2.1], a finite sum of scalar multiples of compact operators is compact, so $\sum_{k = 1}^{m} a_{k} K^{k}$ is compact.

step 1.4step 2.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Norm limit of compact operators is compact

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $X$ be a normed space, let $Y$ be a Banach space (Banach space), and let $T_{n} : X \to Y$ , $n \in N$ , be compact operators (Compact linear operator) with $∥ T_{n} - T ∥ \to 0$ for a bounded linear operator $T : X \to Y$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ). Then $T$ is compact.

Facts & Assumptions

[A1]

$S : X \to Y$ is compact exactly when $\overline{S ({\overline{B}}_{X})}$ is compact, where ${\overline{B}}_{X} = {x \in X : ∥ x ∥ \leq 1}$ (Compact linear operator); and $∥ S x ∥ \leq ∥ S ∥ ∥ x ∥$ for every $x$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces).

[A2]

A compact metric space $(M, d)$ has a finite subcover of every open cover; in particular, if every point of $M$ has a ball of the family ${M \cap B (y, ε) : y \in M}$ containing it, then finitely many of these balls cover $M$ (Open cover, subcover, compact metric space, and compact subset of a metric space, Open ball, closed ball and sphere in a metric space). Selecting one index from each of finitely many nonempty index sets is possible without choice (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

[A3]

For nonempty $A$ in a metric space, $\overline{A} = {y : d (y, A) = 0}$ (claim 1 of The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset); so for $y \in \overline{A}$ and real $δ > 0$ there is $a \in A$ with $d (y, a) < δ$ (Open ball, closed ball and sphere in a metric space).

[A4]

A subset $A$ of a metric space is totally bounded when for every real $ε > 0$ there are finitely many points $f_{0}, \dots, f_{m} \in A$ with $A \subseteq ⋃_{i \leq m} B (f_{i}, ε)$ (Finite $ε$ -net and totally bounded metric space, Open ball, closed ball and sphere in a metric space).

[A5]

A closed subset of a complete metric space is complete (claim 2 of Closed subspaces of complete metric spaces are complete; the converse under countable choice), and a Banach space is a complete normed space (Banach space); a complete and totally bounded metric space is compact under ${A C}_{ω}$ (A complete, totally bounded metric space is compact, proved from countable choice used exactly once).

Proof

technique · direct

Given: ${A C}_{ω}$ , a normed space $X$ , a Banach space $Y$ , compact operators $T_{n} : X \to Y$ with $∥ T_{n} - T ∥ \to 0$ , and $C_{n} : = \overline{T_{n} ({\overline{B}}_{X})}$ .

1.1

For every real $ε > 0$ there is $n$ with $∥ T - T_{n} ∥ < ε$ : this is exactly the convergence $∥ T_{n} - T ∥ \to 0$ in the norm metric of $B (X, Y)$ .

1.2

For every $c \in C_{n}$ and every real $δ > 0$ there is $x \in {\overline{B}}_{X}$ with $∥ c - T_{n} x ∥ < δ$ , because $c \in \overline{T_{n} ({\overline{B}}_{X})}$ , the set $T_{n} ({\overline{B}}_{X})$ is nonempty, and [A3] applies to $A = T_{n} ({\overline{B}}_{X}) \subseteq Y$ .

1.3

Each $C_{n}$ is compact by [A1].

1.4

Whenever $A$ is a totally bounded nonempty subset of a metric space, its closure is totally bounded: given $ε > 0$ , [A4] gives finitely many points $f_{0}, \dots, f_{m} \in A$ with $A \subseteq ⋃_{i \leq m} B (f_{i}, ε / 2)$ ; for $y \in \overline{A}$ there is $a \in A$ with $d (y, a) < ε / 2$ by [A3], and $a \in B (f_{i}, ε / 2)$ for some $i$ , so $d (y, f_{i}) < ε$ ; hence the same finite set, whose points lie in $\overline{A}$ , is an $ε$ -net for $\overline{A}$ .

A3A4algebra

1.5

The set $\overline{T ({\overline{B}}_{X})}$ is closed in the complete space $Y$ , hence a complete metric space by [A5].

2.1

Fix a real $ε > 0$ and choose $n$ with $∥ T - T_{n} ∥ < ε / 4$ by [step 1.1]. The family of balls $C_{n} \cap B (T_{n} x, ε / 4)$ , $x \in {\overline{B}}_{X}$ , covers $C_{n}$ by [step 1.2]; since $C_{n}$ is compact by [step 1.3], finitely many indices $x_{0}, \dots, x_{m} \in {\overline{B}}_{X}$ satisfy $C_{n} \subseteq ⋃_{i \leq m} B (T_{n} x_{i}, ε / 4)$ , and one may pick these finitely many indices by [A2]. For every $x \in {\overline{B}}_{X}$ the point $T_{n} x$ lies in $C_{n}$ , so some $i$ has $∥ T_{n} x - T_{n} x_{i} ∥ < ε / 4$ ; writing $T x - T x_{i} = (T_{n} x - T_{n} x_{i}) + (T - T_{n}) (x - x_{i})$ and using $∥ x ∥, ∥ x_{i} ∥ \leq 1$ gives $∥ T x - T x_{i} ∥ \leq ∥ T_{n} x - T_{n} x_{i} ∥ + ∥ T - T_{n} ∥ ∥ x - x_{i} ∥ < ε / 4 + ε / 2 < ε$ ; thus ${T x_{0}, \dots, T x_{m}}$ is a finite $ε$ -net for $T ({\overline{B}}_{X})$ with centres in $T ({\overline{B}}_{X})$ , and $T ({\overline{B}}_{X})$ is totally bounded.

step 1.1step 1.2step 1.3A1A2algebra

3.1

By [step 1.4] the closure $\overline{T ({\overline{B}}_{X})}$ is totally bounded as well, its finite nets having centres in that closure.

step 1.4step 2.1

4.1

The space $\overline{T ({\overline{B}}_{X})}$ is complete by [step 1.5] and totally bounded by [step 3.1]; by [A5] it is compact, and therefore $T$ is compact by [A1].

step 1.5step 3.1A1A5∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-09-22Open item page →

Approximable operator

Definition

Let $X$ and $Y$ be normed spaces over the same scalar field and let $B (X, Y)$ be the space of bounded linear operators with the operator norm (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators, A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum). Write

$F (X, Y) : = {R \in B (X, Y) : R (X) admits an ordered basis of finite length}$

for the set of bounded finite-rank operators (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis). An operator $T \in B (X, Y)$ is approximable when $T$ lies in the closure of $F (X, Y)$ in the operator-norm metric (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

The closure is an epsilon statement. The zero operator lies in $F (X, Y)$ , since its range ${0}$ admits the empty ordered basis, so $F (X, Y)$ is nonempty and the metric-space description of the closure applies: $T$ is approximable if and only if for every real $ε > 0$ there is a bounded finite-rank operator $R$ with $∥ T - R ∥ < ε$ . No choice principle is used for this equivalence.

Approximable operators are compact, under countable choice. Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), let $Y$ be a Banach space (Banach space) and let $T$ be approximable. Choosing for every $n \in N$ a bounded finite-rank $R_{n}$ with $∥ T - R_{n} ∥ < 1 / (n + 1)$ is a countable selection from nonempty sets, so such operators exist; each $R_{n}$ is compact (Bounded finite rank operators are compact) and $∥ R_{n} - T ∥ \to 0$ , so $T$ is compact by the norm-limit theorem (Norm limit of compact operators is compact, Compact linear operator).

No converse is asserted here. The statement that every compact operator into $Y$ is approximable is the approximation-property question for the target $Y$ ; it is not a consequence of the definition and is not claimed. The companion page records the implication that holds when $Y$ has the approximation property, and the distinction between compact and approximable operators for a general Banach target is left open.

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Schauder compact adjoint theorem

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ and $Y$ be Banach spaces over the same scalar field and let $T : X \to Y$ be a bounded linear operator (A bounded linear operator between normed spaces), with transpose $T^{*} : Y^{*} \to X^{*}$ (The transpose of a bounded operator, The dual space X^* of a normed space and its dual norm). Then $T$ is compact (Compact linear operator) if and only if $T^{*}$ is compact.

Facts & Assumptions

[A1]

$S$ is compact exactly when $\overline{S ({\overline{B}}_{X})}$ is compact (Compact linear operator); the transpose is the bounded linear map $(S^{*} h) (x) = h (S x)$ with $∥ S^{*} ∥ = ∥ S ∥$ (The transpose of a bounded operator, The transpose is bounded with the same norm, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A2]

If $X$ is Banach then $X^{*} = B (X, K)$ is Banach (If (Y) is Banach then (\mathcal B(X,Y)) is Banach, The dual space X^* of a normed space and its dual norm, Every finite-dimensional normed space is Banach), and a closed subset of a Banach space is complete (claim 2 of Closed subspaces of complete metric spaces are complete; the converse under countable choice, Banach space).

[A3]

Assume ${A C}_{ω}$ and $D C$ : compact, sequentially compact and "complete and totally bounded" agree for metric spaces (For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice). Under $A C$ , both hold (AC supplies the countable and dependent choices used in Banach integration, The Axiom of Countable Choice ( ${A C}_{ω}$ ), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[A4]

A compact metric space has a finite subcover of every open cover; the balls centred at points of a nonempty compact set cover it, and choosing one index per member of a finite subcover is choice-free (Open cover, subcover, compact metric space, and compact subset of a metric space, Open ball, closed ball and sphere in a metric space, Every natural-number-indexed list of nonempty sets has a choice function on its family of values). An at most countable union of finite sets is at most countable under ${A C}_{ω}$ (Countable unions of at most countable sets, assuming ${A C}_{ω}$ ).

[A5]

For nonempty $A$ in a metric space, $A \subseteq \overline{A} = {y : d (y, A) = 0}$ (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset); a bounded sequence in $K$ has a convergent subsequence, $K = R$ by Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence and $K = C$ by the isometry $C = R^{2}$ of The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane together with For $n \geq 1$ every bounded sequence in $R^{n}$ has a convergent subsequence.

[A6]

The canonical maps $J_{X} : X \to X^{* *}$ are linear isometries (The canonical bidual map is an isometry) with $S^{* *} J_{X} = J_{Y} S$ for bounded $S$ (The canonical map is natural); an isometric image of a Banach space is a closed subspace (Closed subspaces of complete metric spaces are complete; the converse under countable choice).

[A7]

If $S$ is compact and $C$ is bounded linear, then $C S$ and $S C$ are compact (Compositions with a compact operator are compact).

Proof

technique · direct

Given: $A C$ , Banach spaces $X, Y$ over one scalar field, a bounded linear $S : X \to Y$ , the transpose $S^{*} : Y^{*} \to X^{*}$ , the closed unit balls ${\overline{B}}_{X}, {\overline{B}}_{Y^{*}}$ , and $Θ : = \overline{S ({\overline{B}}_{X})}$ .

1.1

Every bounded sequence in $K$ has a convergent subsequence.

1.2

If $S$ is compact then $Θ$ is compact by [A1], so for every real $δ > 0$ there are finitely many points $x_{0}, \dots, x_{m} \in {\overline{B}}_{X}$ with $Θ \subseteq ⋃_{i \leq m} B (S x_{i}, δ)$ : the balls $Θ \cap B (S x, δ)$ , $x \in {\overline{B}}_{X}$ , cover $Θ$ by [A5], compactness gives a finite subcover, and one index per member of that finite subcover may be chosen by [A4].

A1A4A5

1.3

The dual $X^{*}$ is Banach by [A2], so the set $C : = \overline{S^{*} ({\overline{B}}_{Y^{*}})} \subseteq X^{*}$ is a complete metric space by [A2].

1.4

If $K$ is a compact operator and $M$ is a closed subspace of its target containing $K (X)$ , then the corestriction $K_{0} : X \to M$ is compact: for bounded $E \subseteq X$ the closure of $K_{0} (E)$ in $M$ equals $\overline{K (E)} \cap M$ , a closed subset of the compact set $\overline{K (E)}$ .

1.5

The space $J_{Y} (Y)$ is a closed subspace of $Y^{* *}$ and the inverse of $J_{Y} : Y \to J_{Y} (Y)$ is a bounded isometry, by [A6].

2.1

Under the hypothesis of [step 1.2], the union $D$ of the finite $(1 / (k + 1))$ -nets of $Θ$ obtained from [step 1.2] for $k \in N$ is at most countable and dense in $Θ$ , the nets being chosen together by ${A C}_{ω}$ and their union counted by [A4]; hence there is a surjection $N \to D$ listing $D$ as $(d_{j})$ .

step 1.2A3A4

2.2

If $(g_{n})$ is a sequence in ${\overline{B}}_{Y^{*}}$ and $D = (d_{j})$ is a countable subset of $Y$ , then there are a strictly increasing index map $n : N \to N$ and scalars to which $g_{n_{j}} (d_{k})$ converges for all $j, k$ : for each fixed $d_{k}$ the scalar sequence $g_{n} (d_{k})$ is bounded by $∥ d_{k} ∥$ and has a convergent subsequence by [step 1.1], and the standard diagonal selection of nested subsequences is licensed by $D C$ .

step 1.1A3

3.1

Assume $S$ compact and let $(g_{n})$ be a sequence in ${\overline{B}}_{Y^{*}}$ . With $D$ as in [step 2.1], [step 2.2] gives a subsequence $(g_{n_{j}})$ with $g_{n_{j}} (d)$ convergent for every $d \in D$ . Given a real $ε > 0$ , choose $k$ with $1 / (k + 1) < ε / 4$ and let $F_{k} \subseteq {\overline{B}}_{X}$ be the finite net of [step 1.2] for $δ = 1 / (k + 1)$ ; convergence on the finite set $F_{k}$ gives $J$ with $∣ (g_{n_{j}} - g_{n_{l}}) (S x_{i}) ∣ < ε / 2$ for all $j, l \geq J$ and all $i \leq m$ , and for $x \in {\overline{B}}_{X}$ one has $∥ S x - S x_{i} ∥ < ε / 4$ for some $i$ , so $∣ (S^{*} g_{n_{j}} - S^{*} g_{n_{l}}) (x) ∣ \leq ∣ (g_{n_{j}} - g_{n_{l}}) (S x_{i}) ∣ + ∥ g_{n_{j}} - g_{n_{l}} ∥ ∥ S x - S x_{i} ∥ < ε / 2 + 2 ε / 4 = ε$ ; hence $(S^{*} g_{n_{j}})$ is Cauchy in $X^{*}$ .

step 2.1step 2.2A1A4

4.1

Under the hypothesis of [step 3.1] the Cauchy sequence $(S^{*} g_{n_{j}})$ converges in the complete space $X^{*}$ by [step 1.3], and its limit lies in $C$ because every $S^{*} g_{n_{j}} \in S^{*} ({\overline{B}}_{Y^{*}}) \subseteq C$ and $C$ is closed; so every sequence in $S^{*} ({\overline{B}}_{Y^{*}})$ has a subsequence converging in $C$ .

step 1.3step 3.1

5.1

Under the hypothesis of [step 3.1], every sequence $(y_{n})$ in $C$ has a subsequence converging in $C$ : choosing $g_{n} \in {\overline{B}}_{Y^{*}}$ with $∥ y_{n} - S^{*} g_{n} ∥ < 1 / (n + 1)$ for every $n$ is a countable selection licensed by [A3], and applying [step 4.1] to $(g_{n})$ yields a subsequence with $S^{*} g_{n_{j}} \to y \in C$ , whence $y_{n_{j}} \to y$ .

step 4.1A3A5

6.1

Under the hypothesis of [step 3.1] the space $C$ is sequentially compact by [step 5.1], hence compact by [A3], and then $S^{*}$ is compact by [A1].

step 5.1A1A3

7.1

Suppose now that $T^{*} : Y^{*} \to X^{*}$ is compact. Both $X^{*}$ and $Y^{*}$ are Banach by [A2], so [step 6.1] applied to the bounded linear operator $T^{*}$ between Banach spaces gives that $(T^{*})^{*} = T^{* *}$ is compact; with [step 1.5] and [A6], $T^{* *} J_{X} = J_{Y} T$ , so $T^{* *} J_{X}$ is compact by [A7], and $T = J_{Y}^{- 1} \circ (T^{* *} J_{X})$ is the composite of the corestriction of $T^{* *} J_{X}$ to the closed subspace $J_{Y} (Y)$ — compact by [step 1.4] — with the bounded operator $J_{Y}^{- 1}$ , hence compact by [A7].

step 1.4step 1.5step 6.1A2A6A7

8.1

Conversely, if $T$ is compact then $T^{*}$ is compact by [step 6.1]; and if $T^{*}$ is compact then $T$ is compact by [step 7.1]; this is the asserted equivalence.

step 6.1step 7.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Compact operator sends weakly convergent sequences to norm convergent sequences

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ and $Y$ be Banach spaces over the same scalar field, let $T : X \to Y$ be a compact operator (Compact linear operator) and let $(x_{n})$ be a sequence in $X$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with $x_{n} ⇀ x$ weakly (Weak convergence of nets and sequences). Then $∥ T x_{n} - T x ∥ \to 0$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

Facts & Assumptions

[A1]

Under the Axiom of Dependent Choice, a family of bounded linear operators on a Banach space that is pointwise bounded is norm bounded (Uniform boundedness principle, Banach space); the continuous dual $X^{*}$ is Banach even when $X$ is incomplete (The continuous dual, its completeness, and evaluation), and the canonical map $J_{X} : X \to X^{* *}$ is a linear isometry (The canonical bidual map is an isometry).

[A2]

$x_{n} ⇀ x$ means $f (x_{n}) \to f (x)$ for every $f \in X^{*}$ ; the transpose satisfies $(T^{*} g) (x) = g (T x)$ and $T^{*} g \in X^{*}$ for $g \in Y^{*}$ (Weak convergence of nets and sequences, The transpose of a bounded operator, A bounded linear operator between normed spaces).

[A3]

Assume $D C$ : if a sequence fails to converge to a point then there are a real $ε > 0$ and a strictly increasing index map $j$ with $∥ T x_{n_{j}} - T x ∥ \geq ε$ for all $j$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , A strictly increasing index map satisfies $n_{k} \geq k$ , The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[A4]

Under $D C$ , a compact operator sends bounded sequences to sequences with convergent subsequences (Sequential characterization of compact operators, Compact linear operator).

[A5]

The dual separates points: $u \neq v$ in a normed space gives $g$ with $g (u) \neq g (v)$ (The dual space separates points of a normed space); and $A C$ implies $D C$ (AC supplies the countable and dependent choices used in Banach integration, The Axiom of Choice, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Proof

technique · direct

Given: $A C$ , Banach spaces $X, Y$ over one scalar field, a compact operator $T : X \to Y$ , a sequence $(x_{n})$ in $X$ with $x_{n} ⇀ x$ .

1.1

The sequence $(x_{n})$ is norm bounded: the operators $J_{X} x_{n} : X^{*} \to K$ are pointwise bounded because $f (x_{n}) \to f (x)$ makes $(f (x_{n}))_{n}$ a bounded scalar sequence for each $f \in X^{*}$ , so [A1] and the isometry property give $\sup_{n} ∥ x_{n} ∥ = \sup_{n} ∥ J_{X} x_{n} ∥ < \infty$ .

A1A2A5

1.2

The sequence $(T x_{n})$ converges to $T x$ weakly: for $g \in Y^{*}$ one has $g (T x_{n}) = (T^{*} g) (x_{n}) \to (T^{*} g) (x) = g (T x)$ by [A2].

1.3

If $∥ T x_{n} - T x ∥ \to̸ 0$ , then by [A3] there are $ε > 0$ and a strictly increasing $j$ with $∥ T x_{n_{j}} - T x ∥ \geq ε$ for every $j$ .

2.1

Assume $∥ T x_{n} - T x ∥ \to̸ 0$ and take $ε$ and $(x_{n_{j}})$ as in [step 1.3]. The subsequence $(x_{n_{j}})$ is bounded by [step 1.1], so [A4] gives a further subsequence $(x_{n_{j_{k}}})$ with $T x_{n_{j_{k}}} \to y$ for some $y \in Y$ .

step 1.1step 1.3A4

3.1

For every $g \in Y^{*}$ the scalar sequence $g (T x_{n_{j_{k}}})$ converges to $g (y)$ because $g$ is bounded hence continuous, and to $g (T x)$ by [step 1.2]; hence $g (y) = g (T x)$ for every $g$ , and [A5] gives $y = T x$ .

step 1.2step 2.1A2A5

4.1

But $∥ T x_{n_{j_{k}}} - T x ∥ \geq ε$ for every $k$ by [step 1.3], contradicting $T x_{n_{j_{k}}} \to T x = y$ from [step 3.1].

step 1.3step 3.1

5.1

Hence the assumption $∥ T x_{n} - T x ∥ \to̸ 0$ is false, that is, $∥ T x_{n} - T x ∥ \to 0$ .

step 4.1∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Kernel of identity minus compact is finite dimensional

Statement

Let $X$ be a normed space over $R$ or $C$ and let $K : X \to X$ be a compact operator (Compact linear operator). Then the kernel

$\ker (I - K) = {x \in X : K x = x}$

is a finite-dimensional subspace of $X$ : it admits an ordered basis of finite length (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

Facts & Assumptions

[A1]

$K$ is a bounded linear operator, and a bounded linear operator is continuous (A bounded linear operator between normed spaces, For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent); $K$ is compact, that is, $\overline{K ({\overline{B}}_{X})}$ is compact (Compact linear operator).

[A2]

The normed space $X$ is a metric space with $d (x, y) = ∥ x - y ∥$ , convergence is metric convergence and limits of sequences are unique (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , A sequence in a metric space has at most one limit, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, Real and complex scalar conventions for normed spaces); the closed unit ball is ${\overline{B}}_{X} = {x \in X : ∥ x ∥ \leq 1}$ and is closed (Open ball, closed ball and sphere in a metric space, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed).

[A3]

If the closed unit ball of a normed space is compact, then that space admits an ordered basis of finite length (The closed unit ball is compact if and only if the normed space is finite-dimensional); a closed subset of a compact metric space is a compact metric subspace (A closed subset of a compact metric space is compact). Relative openness is the trace of ambient openness, and compactness is the compactness of the restricted metric (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

Proof

technique · direct

Given: A normed space $X$ over $R$ or $C$ , a compact operator $K : X \to X$ , and $N : = \ker (I - K) = {x \in X : K x = x}$ .

1.1

The set $N$ is a linear subspace of $X$ , since $I - K$ is linear. It is closed without using a sequential-closure criterion: take a bound $C \geq 0$ for $K$ . If $x \notin N$ , put $a = ∥ x - K x ∥ > 0$ . For $∥ y - x ∥ < a / (2 (1 + C))$ , the triangle inequality gives $∥ y - K y ∥ \geq a - ∥ (I - K) (y - x) ∥ \geq a - (1 + C) ∥ y - x ∥ > a / 2 > 0$ . Thus a ball around every $x \notin N$ lies outside $N$ , proving its complement open.

A1A2

2.1

Every $x$ in the closed unit ball $N \cap {\overline{B}}_{X}$ of $N$ satisfies $x = K x$ and $∥ x ∥ \leq 1$ , hence $x \in K ({\overline{B}}_{X})$ ; therefore $N \cap {\overline{B}}_{X} \subseteq K ({\overline{B}}_{X}) \subseteq \overline{K ({\overline{B}}_{X})}$ , and $N \cap {\overline{B}}_{X}$ is a closed subset of $X$ by [step 1.1] and [A2].

step 1.1A2

3.1

The set $\overline{K ({\overline{B}}_{X})}$ is compact by [A1], so by [step 2.1] and [A3] the set $N \cap {\overline{B}}_{X}$ is compact with its metric restricted from $X$ : it is relatively closed in the compact metric subspace $\overline{K ({\overline{B}}_{X})}$ . Restricting that same metric via $N$ gives exactly the same distance on $N \cap {\overline{B}}_{X}$ , so it is also the compact closed unit ball of the normed space $N$ , so $N$ admits an ordered basis of finite length by [A3].

step 2.1A1A3

4.1

Hence $\ker (I - K)$ is finite dimensional, as claimed.

step 3.1∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Range of identity minus compact is closed

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Let $X$ be a Banach space over $R$ or $C$ , let $K : X \to X$ be a compact operator (Compact linear operator) and put $A : = I - K$ . Then $ran A$ is a closed subspace of $X$ , and there is a real $C > 0$ with

$dist (x, \ker A) \leq C ∥ A x ∥ for every x \in X,$

the distance being the quotient seminorm of The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M)).

Facts & Assumptions

[A1]

$ran A$ is closed if and only if there is a real $C > 0$ with $dist (x, \ker A) \leq C ∥ A x ∥$ for every $x$ , under DC for the bounded linear map $A$ between Banach spaces (Closed range is equivalent to a quotient estimate); here $dist (x, M) = ∥ x + M ∥_{X / M}$ is the quotient seminorm (The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))).

[A2]

$K$ is bounded and bounded linear operators are continuous (A bounded linear operator between normed spaces, For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent); DC implies Countable Choice (Dependent choice implies countable choice), and under DC a compact operator maps bounded sequences to sequences with convergent subsequences (Sequential characterization of compact operators, Sequences of reals: bounded, eventually, frequently, tails, subsequences, A strictly increasing index map satisfies $n_{k} \geq k$ ).

[A3]

In a metric space, limits of sequences are unique (A sequence in a metric space has at most one limit, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ); the distance to a fixed set is 1-Lipschitz, $∣ dist (u, N) - dist (v, N) ∣ \leq ∥ u - v ∥$ (The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))).

Proof

technique · direct

Given: $D C$ , a Banach space $X$ over $R$ or $C$ , a compact operator $K : X \to X$ , and $A = I - K$ , $N : = \ker A$ .

1.1

$N$ is a closed linear subspace: $A$ is bounded hence continuous by [A2], so $x_{j} \in N$ with $x_{j} \to x$ gives $A x_{j} = 0 \to A x$ , whence $A x = 0$ by [A3].

A2A3

1.2

If the estimate of [A1] fails for every $C$ , then for each fixed $n \in N$ the set $W_{n} : = {w : dist (w, N) = 1, ∥ w ∥ \leq 2, ∥ A w ∥ < 1 / (n + 1)}$ is nonempty: taking $C = n + 1$ gives $x$ with $dist (x, N) > (n + 1) ∥ A x ∥ \geq 0$ , so the distance is positive; choose $m \in N$ with $∥ x - m ∥ < 2 dist (x, N)$ by the definition of the infimum, and set $w : = (x - m) / dist (x, N)$ . Scaling the distance gives $dist (w, N) = 1$ , the norm bound $∥ w ∥ < 2$ holds, and $∥ A w ∥ = ∥ A x ∥ / dist (x, N) < 1 / (n + 1)$ because $A m = 0$ .

step 1.1A1A3algebra

2.1

Assume the estimate fails. By Countable Choice in [A2], select $w_{n} \in W_{n}$ for every $n$ as in [step 1.2], and put $M : = {x \in X : ∥ x ∥ \leq 2}$ . The sequence $(w_{n})$ lies in the bounded set $M$ , and $K$ is compact, so by [A2] there is a strictly increasing $j$ with $K w_{n_{j}} \to z$ for some $z \in X$ .

step 1.2A2

3.1

Along that subsequence, $w_{n_{j}} = A w_{n_{j}} + K w_{n_{j}} \to z$ , because $∥ A w_{n_{j}} ∥ < 1 / (n_{j} + 1) \to 0$ by [step 1.2] and $K w_{n_{j}} \to z$ by [step 2.1].

step 1.2step 2.1algebra

4.1

The limit $z$ lies in $N$ : by [step 3.1] and the continuity of $A$ , $A z = \lim_{j} A w_{n_{j}} = 0$ .

step 3.1A2A3

5.1

But this contradicts $dist (w_{n_{j}}, N) = 1$ : by [A3] the numbers $dist (w_{n_{j}}, N)$ converge to $dist (z, N)$ , so $dist (z, N) = 1$ , whereas $z \in N$ forces $dist (z, N) = 0$ .

step 1.2step 3.1step 4.1A3

6.1

Hence the estimate of [A1] holds for some real $C > 0$ , and then [A1] gives that $ran A$ is closed.

step 5.1A1∎

LemmaStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Riesz Schauder ascent and descent stabilize

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Let $X$ be a Banach space over $R$ or $C$ , let $K : X \to X$ be a compact operator (Compact linear operator) and put $A : = I - K$ . Then there is $m_{0} \in N$ such that for every $n \geq m_{0}$

$\ker A^{n} = \ker A^{m_{0}}, ran A^{n} = ran A^{m_{0}},$

and for every such $m$ the following hold:

$X = \ker A^{m} \oplus ran A^{m}$ as a direct sum of linear subspaces;
$\ker A^{m}$ is finite dimensional (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis) and $ran A^{m}$ is closed in $X$ ;
$A$ maps $ran A^{m}$ bijectively onto itself, and the restricted map $ran A^{m} \to ran A^{m}$ , $y \mapsto A y$ , is a bounded linear isomorphism with bounded inverse (Bounded inverse theorem).

Facts & Assumptions

[A1]

For every $n \geq 1$ there is a compact $K_{n}$ with $A^{n} = I - K_{n}$ : $A = I - K$ , and if $A^{n} = I - K_{n}$ with $K_{n}$ compact then $A^{n + 1} = (I - K_{n}) A = I - K_{n} - K + K_{n} K$ , where $K_{n} + K - K_{n} K$ is compact by the ideal and linear-subspace properties (Compositions with a compact operator are compact, Linear combinations of compact operators are compact, Compact linear operator, A bounded linear operator between normed spaces).

[A2]

For compact $C$ the kernel $\ker (I - C)$ is finite dimensional (Kernel of identity minus compact is finite dimensional) and $ran (I - C)$ is closed, with $dist (x, \ker (I - C)) \leq C_{0} ∥ (I - C) x ∥$ for some real $C_{0} > 0$ (Range of identity minus compact is closed, The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))).

[A3]

Riesz lemma: for a proper closed subspace $M$ of a normed space and $0 < α < 1$ there is $x$ with $∥ x ∥ = 1$ and $dist (x, M) > α$ (Riesz lemma, The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M)), A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[A4]

Under DC, Countable Choice is available, and a compact operator sends bounded sequences to sequences with convergent subsequences (Dependent choice implies countable choice, Sequential characterization of compact operators, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ); a subsequence of a bounded sequence is bounded.

[A5]

A closed linear subspace of a Banach space is a Banach space (A closed subspace of a Banach space is Banach), and a bounded bijection between Banach spaces has a bounded inverse (Bounded inverse theorem, Banach space).

Proof

technique · direct

Given: $D C$ , a Banach space $X$ over $R$ or $C$ , a compact $K : X \to X$ , and $A = I - K$ .

1.1

For every $n \geq 1$ the operator $A^{n}$ equals $I - K_{n}$ with $K_{n}$ compact.

2.1

For every $n \geq 1$ the kernel $\ker A^{n}$ is finite dimensional with $\ker A^{n} \subseteq \ker A^{n + 1}$ , and the range $ran A^{n}$ is closed with $ran A^{n + 1} \subseteq ran A^{n}$ .

step 1.1A2

3.1

For every $n$ with $\ker A^{n} \neq \ker A^{n + 1}$ , the finite-dimensional subspace $\ker A^{n}$ is closed in the normed space $\ker A^{n + 1}$ , so Riesz's lemma gives $x \in \ker A^{n + 1}$ with $∥ x ∥ = 1$ and $dist (x, \ker A^{n}) > 1 / 2$ (A finite-dimensional normed subspace is closed).

step 2.1A3

3.2

For every $n$ with $ran A^{n} \neq ran A^{n + 1}$ there is $x \in ran A^{n}$ with $∥ x ∥ = 1$ and $dist (x, ran A^{n + 1}) > 1 / 2$ , the distance being computed in $X$ .

step 2.1A3

4.1

The chain $\ker A^{n}$ stabilizes: if $\ker A^{n} \neq \ker A^{n + 1}$ for every $n$ , Countable Choice in [A4] applied to [step 3.1] gives unit vectors $x_{n} \in \ker A^{n + 1}$ with $dist (x_{n}, \ker A^{n}) > 1 / 2$ ; for $m < n$ one has $x_{m} \in \ker A^{n}$ , $A x_{m} \in \ker A^{n}$ and $A x_{n} \in \ker A^{n}$ , so $K x_{n} - K x_{m} = x_{n} - (x_{m} + A x_{n} - A x_{m})$ is at distance $> 1 / 2$ from $0$ , and the bounded sequence $(K x_{n})$ has no convergent subsequence, contradicting [A4].

step 3.1A4

4.2

The chain $ran A^{n}$ stabilizes: if $ran A^{n} \neq ran A^{n + 1}$ for every $n$ , Countable Choice in [A4] applied to [step 3.2] gives unit vectors $x_{n} \in ran A^{n}$ with $dist (x_{n}, ran A^{n + 1}) > 1 / 2$ ; for $m > n$ one has $x_{m} \in ran A^{n + 1}$ , $A x_{n} \in ran A^{n + 1}$ and $A x_{m} \in ran A^{n + 1}$ , so $K x_{n} - K x_{m} = x_{n} - (x_{m} + A x_{n} - A x_{m})$ has norm $> 1 / 2$ , and the bounded sequence $(K x_{n})$ has no convergent subsequence, contradicting [A4].

step 3.2A4

5.1

Choose $m$ so that both chains are constant from $m$ onward by [step 4.1] and [step 4.2], and put $N : = \ker A^{m}$ , $Y : = ran A^{m}$ . Then $X = N + Y$ : for $x \in X$ the element $A^{m} x$ lies in $Y = ran A^{2 m}$ , so $A^{m} x = A^{2 m} y$ for some $y \in X$ , and $x - A^{m} y \in N$ .

step 4.1step 4.2

6.1

Under the choice of [step 5.1] one has $N \cap Y = {0}$ : if $x \in N \cap Y$ , say $x = A^{m} y$ with $A^{m} x = 0$ , then $A^{2 m} y = 0$ , so $y \in \ker A^{2 m} = \ker A^{m}$ by [step 4.1] and $x = A^{m} y = 0$ .

step 5.1

6.2

Under the choice of [step 5.1], $A (Y) = Y$ : $A (Y) = A (A^{m} X) = A^{m + 1} X = Y$ by the stabilization of [step 4.2].

step 5.1

7.1

Under the choice of [step 5.1], $X = N \oplus Y$ , and $N$ is finite dimensional and $Y$ is closed by [step 2.1].

step 5.1step 6.1step 2.1

8.1

Under the choice of [step 5.1], the restriction $A ∣_{Y}$ is injective: if $A y = 0$ with $y = A^{m} z \in Y$ , then $A^{m + 1} z = 0$ , so $z \in \ker A^{m + 1} = \ker A^{m}$ by [step 4.1] and $y = A^{m} z = 0$ .

step 7.1

9.1

Under the choice of [step 5.1], $A ∣_{Y} : Y \to Y$ is a bounded bijection by [step 6.2] and [step 8.1], and $Y$ is a Banach space by [step 7.1] and [A5], so the inverse $(A ∣_{Y})^{- 1}$ is bounded by [A5].

step 7.1step 6.2step 8.1A5

10.1

Taking $m_{0} : = m$ from [step 5.1] gives the stabilization, and [step 7.1], [step 6.2] and [step 9.1] give the decomposition, the finite-dimensional kernel, the closed range and the bounded isomorphism on that range; every larger $m$ works as well because the chains are constant from $m_{0}$ onward.

step 5.1step 7.1step 6.2step 9.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Fredholm alternative for identity minus compact

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ be a Banach space over a fixed field $F \in {R, C}$ , let $K : X \to X$ be a compact operator (Compact linear operator) and put $A : = I - K$ . Then:

$A$ is injective if and only if it is surjective, and in that case $A$ is boundedly invertible;
for every $y \in X$ the equation $A x = y$ has a solution if and only if $φ (y) = 0$ for every $φ \in \ker A^{*}$ , the transpose $A^{*} = I - K^{*}$ acting on $X^{*}$ (The transpose of a bounded operator);
$\ker A$ and the cokernel $X / ran A$ (The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M)) are finite dimensional with equal dimensions over $F$ (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

Facts & Assumptions

[A1]

The compact $K$ is bounded; if $C$ is a bound for $K$ , then $∥ A x ∥ \leq (1 + C) ∥ x ∥$ , so $A$ is bounded and linear (Compact linear operator, A bounded linear operator between normed spaces). Under DC there is $m_{0}$ from which the kernel and range chains stabilize; since every larger exponent has the same properties, take $m \geq \max {m_{0}, 1}$ . Then $\ker A^{m} = \ker A^{n}$ and $ran A^{m} = ran A^{n}$ for all $n \geq m$ , and with $N : = \ker A^{m}$ , $Y : = ran A^{m}$ one has $X = N \oplus Y$ , $N$ finite dimensional, $Y$ closed, $A (Y) = Y$ and $A ∣_{Y}$ a bounded isomorphism (Riesz Schauder ascent and descent stabilize, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain); $A C$ supplies DC (AC supplies the countable and dependent choices used in Banach integration).

[A2]

$A^{*} = I - K^{*}$ : the transpose of the identity is the identity and the transpose is additive (Transposition reverses composition, The transpose of a bounded operator); under DC the range of $I - K$ is closed (Range of identity minus compact is closed). For a bounded linear $T$ with closed range one has $(ran T)^{⊥} = \ker T^{*}$ and $\overline{ran T} =^{⊥} (\ker T^{*})$ (Elementary kernel and range annihilator identities).

[A3]

A bounded bijection between Banach spaces has a bounded inverse (Bounded inverse theorem, Banach space). For a linear map $T : V \to W$ over $F$ with $V$ finite dimensional, $\dim_{F} V = \dim_{F} \ker T + \dim_{F} ran T$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ , Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis); for a linear subspace $U \leq V$ the quotient $V / U$ is a vector space and a surjective linear map induces a linear isomorphism $V / \ker T ≅ ran T$ (First isomorphism theorem for vector spaces: $V / \ker T$ is isomorphic to $im T$ , The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M), Linear subspace of a vector space).

Proof

technique · direct

Given: $A C$ , a Banach space $X$ over the fixed field $F \in {R, C}$ , a compact $K : X \to X$ , $A = I - K$ ; and $m$ , $N = \ker A^{m}$ , $Y = ran A^{m}$ as in [A1].

1.1

$N$ is finite dimensional, $Y$ is closed, $A ∣_{Y}$ is a bounded isomorphism of $Y$ onto $Y$ , and $X = N \oplus Y$ . Moreover $A (N) \subseteq N$ , because $A^{m} (A n) = A (A^{m} n) = 0$ for $n \in N$ . All subspaces, quotients and dimensions below are over $F$ .

2.1

$\ker A \subseteq N$ and $\ker A = \ker (A ∣_{N})$ : if $A x = 0$ then $A^{m} x = 0$ , so $x \in N$ ; conversely $x \in N$ with $A x = 0$ means $x \in \ker A$ .

step 1.1

2.2

$ran A = A (N) + Y$ : because $A$ is linear, $X = N + Y$ and $A (Y) = Y$ .

step 1.1

2.3

$A (N) \neq N$ whenever $N \neq {0}$ : $A^{m}$ vanishes on $N$ , and if $A (N) = N$ then $A ∣_{N}$ is a surjective linear endomorphism of the finite-dimensional space $N$ , hence also injective, so $A^{m} ∣_{N} = A ∣_{N}^{m}$ would be injective while $A^{m} ∣_{N} = 0$ and $N \neq {0}$ , a contradiction.

step 1.1A3

3.1

$A^{*} = I - K^{*}$ , and $ran A$ is closed by [A2], so $(ran A)^{⊥} = \ker A^{*}$ and $\overline{ran A} =^{⊥} (\ker A^{*})$ with the closure equal to $ran A$ itself.

step 2.2A2

3.2

The inclusion $N ↪ X$ induces a linear isomorphism $N / A (N) ≅ X / ran A$ : the map $φ (n) = n + ran A$ has kernel $N \cap ran A = N \cap (A (N) + Y) = A (N) + (N \cap Y) = A (N)$ , and it is surjective because every coset $x + ran A$ with $x = n + y$ equals $n + ran A$ .

step 1.1step 2.2A3

3.3

The following are equivalent: $A$ injective, $\ker A = {0}$ , $N = {0}$ , $A$ surjective. Indeed $\ker A = {0}$ is $A$ injective; $N = {0}$ gives $\ker A = {0}$ by [step 2.1], and conversely $N \neq {0}$ makes $A ∣_{N}$ non-injective by [step 2.3], so $\ker A \neq {0}$ ; finally $N = {0}$ gives $X = Y = ran A$ so $A$ is surjective, while if $N \neq {0}$ and $A$ is surjective then $N = N \cap ran A = A (N) + (N \cap Y) = A (N)$ , contradicting [step 2.3].

step 1.1step 2.1step 2.2step 2.3A3

4.1

If $A$ is injective, hence bijective by [step 3.3], then $A^{- 1}$ is bounded by [A3].

step 3.3A3

4.2

For $y \in X$ the equation $A x = y$ is solvable if and only if $y \in ran A$ if and only if $φ (y) = 0$ for every $φ \in \ker A^{*}$ , by [step 3.1].

step 3.1A2

4.3

Rank–nullity for $A ∣_{N} : N \to N$ gives $\dim_{F} N = \dim_{F} \ker A + \dim_{F} A (N)$ . The surjective quotient map $N \to N / A (N)$ has kernel $A (N)$ , so rank–nullity also gives $\dim_{F} N = \dim_{F} A (N) + \dim_{F} (N / A (N))$ . Both its image and the first map's kernel are finite dimensional by rank–nullity. Cancelling the common natural summand and using the isomorphism in step 3.2 yields $\dim_{F} \ker A = \dim_{F} (X / ran A)$ , with both spaces finite dimensional.

step 2.1step 3.2A3algebra

5.1

Collecting: [step 3.3] and [step 4.1] give claim 1, [step 4.2] gives claim 2, and [step 4.3] gives finite dimensionality and equality of the dimensions of kernel and cokernel, claim 3.

step 3.3step 4.1step 4.2step 4.3∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Neumann series and small perturbations of bounded inverses

Statement

Let $X$ and $Y$ be Banach spaces over the same scalar field (Banach space).

If $R \in B (X)$ satisfies $∥ R ∥ < 1$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum), then $I - R$ is invertible with inverse the operator-norm limit of the partial sums $\sum_{n < N} R^{n}$ , the norm limit being taken in $B (X)$ (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators), and $∥ \sum_{n = 0}^{\infty} R^{n} ∥ \leq \frac{1}{1 - ∥ R ∥} .$
If $A \in B (X, Y)$ is invertible with $A^{- 1} \in B (Y, X)$ and $E \in B (X, Y)$ satisfies $∥ A^{- 1} E ∥ < 1$ , then $A + E$ is invertible with $(A + E)^{- 1} = (I + A^{- 1} E)^{- 1} A^{- 1} \in B (Y, X)$ .

Facts & Assumptions

[A1]

$∥ R^{n} ∥ \leq ∥ R ∥^{n}$ for every $n$ , by induction from $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ (Composition satisfies |ST|\le|S|,|T|, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A2]

For $∣ r ∣ < 1$ the scalar series $\sum r^{k}$ converges with sum $1 / (1 - r)$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges); in particular $∥ R ∥ < 1$ makes $\sum_{n} ∥ R ∥^{n}$ converge to the real number $1 / (1 - ∥ R ∥)$ .

[A3]

If $Y$ is Banach then $B (X, Y)$ is Banach for the operator norm (If (Y) is Banach then (\mathcal B(X,Y)) is Banach, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators, Banach space); a series in a Banach space that converges absolutely converges (Series criterion for Banach spaces).

[A4]

Addition and scalar multiplication are continuous on a normed space (Vector addition and scalar multiplication are continuous in a normed space), and the reverse triangle inequality makes every norm continuous with respect to norm convergence (The reverse triangle inequality in a normed space, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

Proof

technique · direct

Given: Banach spaces $X, Y$ over one scalar field, $R \in B (X)$ with $∥ R ∥ < 1$ , and the partial sums $S_{N} : = \sum_{n < N} R^{n}$ .

1.1

For every $n$ one has $∥ R^{n} ∥ \leq ∥ R ∥^{n}$ , and $\sum_{n} ∥ R ∥^{n}$ converges to $1 / (1 - ∥ R ∥)$ .

A1A2

2.1

The space $B (X)$ is Banach, so the absolutely convergent series $\sum_{n} R^{n}$ converges in operator norm to some $S \in B (X)$ . For every $N$ , the finite triangle inequality and [step 1.1] give $∥ S_{N} ∥ \leq \sum_{n < N} ∥ R^{n} ∥ \leq \sum_{n = 0}^{\infty} ∥ R ∥^{n} = \frac{1}{1 - ∥ R ∥} .$ Since $S_{N} \to S$ and the norm is continuous, taking the limit yields $∥ S ∥ \leq 1 / (1 - ∥ R ∥)$ .

step 1.1A2A3A4

2.2

For every $N$ one has $(I - R) S_{N} = S_{N} (I - R) = I - R^{N}$ , and $∥ R^{N} ∥ \leq ∥ R ∥^{N} \to 0$ , so $R^{N} \to 0$ .

step 1.1A1algebra

3.1

From [step 2.1] and [step 2.2], $(I - R) S = \lim_{N} (I - R) S_{N} = \lim_{N} (I - R^{N}) = I$ and likewise $S (I - R) = I$ : the first limit holds because $∥ (I - R) (S - S_{N}) ∥ \leq (1 + ∥ R ∥) ∥ S - S_{N} ∥ \to 0$ .

step 2.1step 2.2A1A4

4.1

Hence $I - R$ is invertible with inverse $S = \sum_{n \geq 0} R^{n}$ and $∥ (I - R)^{- 1} ∥ \leq 1 / (1 - ∥ R ∥)$ , which is claim 1.

step 2.1step 3.1

4.2

For the perturbation, $A + E = A (I + A^{- 1} E)$ and $∥ - A^{- 1} E ∥ = ∥ A^{- 1} E ∥ < 1$ , so by [step 3.1] applied to $R : = - A^{- 1} E \in B (X)$ the operator $I + A^{- 1} E$ is invertible with bounded inverse, and therefore $(A + E)^{- 1} = (I + A^{- 1} E)^{- 1} A^{- 1} \in B (Y, X)$ , which is claim 2.

step 3.1algebra

5.1

Claims 1 and 2 are exactly the two parts of the statement.

step 4.1step 4.2∎

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Spectrum and resolvent of a bounded operator

Definition

Let $X$ be a complex Banach space (Banach space, The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane) and let $T \in B (X)$ be a bounded linear operator (A bounded linear operator between normed spaces, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators). For a scalar $λ \in C$ write $λ I - T$ for the bounded linear operator $x \mapsto λ x - T x$ (Linear map between vector spaces over the same field).

The resolvent set of $T$ is $ρ (T) : = {λ \in C : λ I - T is bijective and its inverse is a bounded operator X \to X} .$ For $λ \in ρ (T)$ the inverse $R (λ, T) : = (λ I - T)^{- 1}$ , an element of $B (X)$ , is the resolvent operator of $T$ at $λ$ .
The spectrum of $T$ is its complement $σ (T) : = C ∖ ρ (T) .$
A scalar $λ$ is an eigenvalue of $T$ when $\ker (λ I - T) \neq {0}$ ; the nonzero vectors of that kernel are the eigenvectors of $T$ for $λ$ , and $E_{λ} (T) : = \ker (λ I - T)$ is the eigenspace. The set of eigenvalues is the point spectrum of $T$ ; plainly the point spectrum is contained in $σ (T)$ , since an operator with nonzero kernel is not injective.
For $λ \in C$ the generalized eigenspace of $T$ at $λ$ is $G_{λ} (T) : = ⋃_{n \geq 1} \ker ((T - λ I)^{n}),$ an increasing union of linear subspaces (Linear subspace of a vector space); the union is a linear subspace because the union is increasing. If $λ$ is an eigenvalue and $G_{λ} (T)$ is finite dimensional, $\dim_{C} G_{λ} (T)$ (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis) is the algebraic multiplicity of the eigenvalue $λ$ .

Every element of $G_{λ} (T) ∖ {0}$ produces an eigenvector, and conversely. If $(T - λ I)^{n} x = 0$ with $x \neq 0$ , let $k \geq 1$ be least with $(T - λ I)^{k} x = 0$ ; then $y : = (T - λ I)^{k - 1} x$ is nonzero and satisfies $(T - λ I) y = 0$ . So $G_{λ} (T) \neq {0}$ exactly when $λ$ is an eigenvalue. In particular $λ \in ρ (T)$ implies $G_{λ} (T) = {0}$ . If $λ$ is an eigenvalue and $G_{λ} (T)$ is finite dimensional, then the algebraic multiplicity is defined and is at least $\dim_{C} E_{λ} (T)$ , because $E_{λ} (T) \subseteq G_{λ} (T)$ .

Conventions. Only complex scalars are treated here; the real case is handled by complexification on a later page, so no spectrum is attached here to a bounded operator on a real Banach space. The definition is purely one of vocabulary; it asserts no nonemptiness of $σ (T)$ , no openness of $ρ (T)$ and no continuity of $λ \mapsto R (λ, T)$ , all of which are proved separately. Since $0 I - T = - T$ , the number $0$ lies in $σ (T)$ exactly when $T$ is not invertible with bounded inverse.

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Riesz schauder spectrum of a compact operator

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ be a complex Banach space, let $K : X \to X$ be a compact operator (Compact linear operator) and let $σ (K)$ be its spectrum (Spectrum and resolvent of a bounded operator). Then:

every $λ \in σ (K)$ with $λ \neq 0$ is an eigenvalue of $K$ whose generalized eigenspace $G_{λ} (K)$ is finite dimensional, so its algebraic multiplicity is finite;
for every real $ε > 0$ the set ${λ \in σ (K) : ∣ λ ∣ \geq ε}$ is finite;
if $X$ is infinite dimensional (does not admit an ordered basis of finite length), then $0 \in σ (K)$ .

Facts & Assumptions

[A1]

For $λ \neq 0$ put $A_{λ} : = I - K / λ$ . Since $K / λ$ is compact, the Fredholm alternative applies to it: $A_{λ}$ is injective if and only if it is surjective, and then boundedly invertible; moreover $λ I - K = λ A_{λ}$ and, for $y \in X$ , the equation $(λ I - K) x = y$ is solvable exactly when $φ (y) = 0$ for every $φ$ in the kernel of the transpose (Linear combinations of compact operators are compact, Fredholm alternative for identity minus compact, Spectrum and resolvent of a bounded operator, A bounded linear operator between normed spaces).

[A2]

For the compact operator $K / λ$ the stabilization lemma gives an $m$ with $\ker A_{λ}^{n} = \ker A_{λ}^{m}$ and $ran A_{λ}^{n} = ran A_{λ}^{m}$ for all $n \geq m$ , $X = \ker A_{λ}^{m} \oplus ran A_{λ}^{m}$ , finite-dimensional $\ker A_{λ}^{m}$ and a bounded isomorphism $A_{λ}$ of $ran A_{λ}^{m}$ onto itself (Riesz Schauder ascent and descent stabilize, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain); AC supplies DC (AC supplies the countable and dependent choices used in Banach integration).

[A3]

If $B \in B (X)$ is invertible with bounded inverse and $∥ (μ - λ) B^{- 1} ∥ < 1$ then $B + (μ - λ) I$ is invertible with bounded inverse; if $∥ C ∥ < 1$ then $I - C$ is invertible with bounded inverse (Neumann series and small perturbations of bounded inverses); and for a nilpotent endomorphism $N$ of a vector space with $N^{m} = 0$ the operator $I + t N$ is invertible with inverse $\sum_{j < m} (- t N)^{j}$ for every scalar $t$ (finite telescoping sum).

[A4]

Under the identification $C = R^{2}$ the metric of $C$ is the Euclidean metric of $R^{2}$ (The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane); a subset of $R^{2}$ is compact exactly when it is closed and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), and every metric open ball is open (Open ball, closed ball and sphere in a metric space, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed).

[A5]

If $T$ is compact and $C$ is bounded linear then $T C$ and $C T$ are compact (Compositions with a compact operator are compact); a normed space whose closed unit ball is compact admits an ordered basis of finite length (The closed unit ball is compact if and only if the normed space is finite-dimensional); a closed subspace of a Banach space is Banach (Closed subspaces of complete metric spaces are complete; the converse under countable choice, Banach space).

Proof

technique · direct

Given: $A C$ , a complex Banach space $X$ , a compact $K : X \to X$ , and for $λ \neq 0$ the operator $A_{λ} = I - K / λ$ .

1.1

For $λ \neq 0$ , $λ \notin σ (K)$ if and only if $\ker (λ I - K) = {0}$ : the operator $λ I - K$ is bijective exactly when it is injective, by the injective-iff-surjective part of [A1] applied to $A_{λ}$ .

1.2

For $λ \neq 0$ the generalized eigenspace satisfies $G_{λ} (K) = ⋃_{n \geq 1} \ker (A_{λ}^{n}) = \ker (A_{λ}^{m})$ for an $m$ given by [A2], hence $G_{λ} (K)$ is finite dimensional whenever it is nonzero, and it is nonzero exactly when $λ$ is an eigenvalue.

A1A2

1.3

The resolvent set $ρ (K)$ is open: if $λ \in ρ (K)$ and $B : = λ I - K$ with inverse $B^{- 1}$ , then for every scalar $μ$ with $∣ μ - λ ∣ ∥ B^{- 1} ∥ < 1$ the operator $μ I - K = B + (μ - λ) I$ is invertible with bounded inverse by [A3], so such $μ$ lie in $ρ (K)$ .

1.4

$σ (K) \subseteq {λ : ∣ λ ∣ \leq ∥ K ∥}$ : if $∣ λ ∣ > ∥ K ∥$ then $∥ K / λ ∥ < 1$ and $λ I - K = λ (I - K / λ)$ is invertible with bounded inverse by [A3], so $λ \in ρ (K)$ .

1.5

If $X$ is infinite dimensional then $0 \in σ (K)$ : if $0 \in ρ (K)$ then $- K = 0 I - K$ has a bounded inverse $S$ , and $I = (- K) S = K (- S)$ is compact by [A5]; then the closed unit ball of $X$ , the image of itself under $I$ , is compact, so $X$ admits an ordered basis of finite length by [A5], a contradiction.

2.1

If $λ \in σ (K)$ and $λ \neq 0$ , then with $N : = G_{λ} (K) = \ker (A_{λ}^{m})$ and $Y : = ran A_{λ}^{m}$ from [A2] one has $X = N \oplus Y$ , $A_{λ}^{m}$ vanishes on $N$ , and $λ I - K$ restricted to $Y$ is invertible with bounded inverse; for every scalar $μ \neq λ$ the operator $μ I - K$ is invertible on $N$ , because on $N$ it equals $(μ - λ) (I + (λ / (μ - λ)) A_{λ})$ and $A_{λ} ∣_{N}$ is nilpotent.

step 1.2A2A3

2.2

Claim 1: if $λ \in σ (K)$ and $λ \neq 0$ , then $λ$ is an eigenvalue with finite-dimensional generalized eigenspace: by [step 1.1] the kernel $\ker (λ I - K)$ is nonzero, and by [step 1.2] the generalized eigenspace is finite dimensional.

step 1.1step 1.2

2.3

For real $ε > 0$ the set $S_{ε} : = {λ \in σ (K) : ∣ λ ∣ \geq ε}$ is compact in $C$ : it is bounded by [step 1.4] and closed because $ρ (K)$ is open by [step 1.3], so under $C = R^{2}$ it is a closed and bounded subset of $R^{2}$ , hence compact by [A4].

step 1.3step 1.4A4

3.1

If $λ \in σ (K)$ and $λ \neq 0$ , there is a real $δ > 0$ with $μ \in ρ (K)$ for every $μ$ with $0 < ∣ μ - λ ∣ < δ$ : choose $δ > 0$ with $δ ∥ ((λ I - K) ∣_{Y})^{- 1} ∥ < 1$ for the $Y$ of [step 2.1]; then for such $μ$ the restriction of $μ I - K$ to $Y$ is invertible by [A3] and its restriction to $N$ is invertible by [step 2.1], and invertibility on both summands of $X = N \oplus Y$ gives invertibility on $X$ .

step 2.1A2A3

4.1

Claim 2: $S_{ε}$ is finite. Every point of $S_{ε}$ lies in $σ (K)$ and is nonzero, so by [step 3.1] each $λ \in S_{ε}$ has a ball $B (λ, δ_{λ})$ meeting $σ (K)$ only in $λ$ ; the sets $S_{ε} \cap B (λ, δ_{λ})$ , $λ \in S_{ε}$ , form an open cover of the compact set $S_{ε}$ by [step 2.3], and each member contains only the single point $λ$ , so a finite subcover exhibits $S_{ε}$ as a finite set.

step 3.1step 2.3A4

5.1

Claim 3 is [step 1.5], and claims 1, 2, 3 are respectively [step 2.2], [step 4.1] and [step 1.5]; the statement is proved.

step 1.5step 2.2step 4.1∎

CorollaryStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Spectrum of a compact operator is countable with only zero as possible accumulation

Statement

$σ (K)$ is at most countable (Finite, countably infinite, countable, uncountable);
for every $λ \in C$ with $λ \neq 0$ there is a real $r > 0$ with $B (λ, r) \cap σ (K) \subseteq {λ}$ (Open ball, closed ball and sphere in a metric space).

In particular the only point of $C$ that can be an accumulation point of $σ (K)$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space) is $0$ .

Facts & Assumptions

[A1]

Under AC, for every real $ε > 0$ the set $S_{ε} : = {λ \in σ (K) : ∣ λ ∣ \geq ε}$ is finite (Riesz schauder spectrum of a compact operator).

[A2]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[A3]

Under ${A C}_{ω}$ , an at most countable union of at most countable sets is at most countable (Countable unions of at most countable sets, assuming ${A C}_{ω}$ , The Axiom of Countable Choice ( ${A C}_{ω}$ ), Finite, countably infinite, countable, uncountable), and every subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable); $A C$ implies ${A C}_{ω}$ (AC supplies the countable and dependent choices used in Banach integration, The Axiom of Choice).

[A4]

In $C$ the balls are those of the metric $d (z, w) = ∣ z - w ∣$ (The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane, Open ball, closed ball and sphere in a metric space); a point $λ$ is an accumulation point of a set $A$ when every punctured ball $B (λ, r) ∖ {λ}$ meets $A$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

Proof

technique · direct

Given: $A C$ , a complex Banach space $X$ , a compact $K : X \to X$ , and the sets $S_{1 / n} = {λ \in σ (K) : ∣ λ ∣ \geq 1 / n}$ , $n \geq 1$ .

1.1

Every subspace of $σ (K)$ of the form $S_{1 / n}$ is finite by [A1]; and every $μ \in σ (K)$ with $μ \neq 0$ lies in some $S_{1 / n}$ , because $∣ μ ∣ > 0$ and [A2] gives $n$ with $1 / n < ∣ μ ∣$ .

A1A2

1.2

The set $U : = {0} \cup ⋃_{n \geq 1} S_{1 / n}$ is at most countable: it is a countable union of finite sets, and [A3] applies.

A1A3

2.1

$σ (K) \subseteq U$ by [step 1.1].

step 1.1

2.2

Claim 2: let $λ \neq 0$ and choose $n$ with $1 / n \leq ∣ λ ∣ / 2$ by [A2]. Every $z \in B (λ, ∣ λ ∣ / 2)$ satisfies $∣ z ∣ \geq ∣ λ ∣ - ∣ z - λ ∣ > ∣ λ ∣ / 2 \geq 1 / n$ , so the set $F : = σ (K) \cap B (λ, ∣ λ ∣ / 2)$ is contained in $S_{1 / n}$ and is finite by [step 1.1]. Put $E : = F ∖ {λ}$ . If $E$ is nonempty, the finite set of positive numbers ${∣ z - λ ∣ : z \in E}$ has a minimum $ρ > 0$ ; if $E$ is empty, put $ρ : = ∣ λ ∣ / 2$ . For $r : = \min (ρ, ∣ λ ∣ / 2) > 0$ , any $z \in B (λ, r) \cap σ (K)$ lies in $F$ , while $z \neq λ$ would put $z$ in $E$ and give the contradiction $∣ z - λ ∣ \geq ρ \geq r > ∣ z - λ ∣$ . Hence $B (λ, r) \cap σ (K) \subseteq {λ}$ .

step 1.1A2A4

3.1

Claim 1: $σ (K)$ is at most countable, being a subset of the at most countable set $U$ , by [A3]; moreover $U$ is at most countable by [step 1.2] and $σ (K) \subseteq U$ by [step 2.1].

step 1.2step 2.1A3

4.1

Finally, if $λ \neq 0$ and every punctured ball around $λ$ met $σ (K)$ , then by [step 2.2] the punctured ball $B (λ, r) ∖ {λ}$ meets $σ (K)$ yet contains none of its points, a contradiction; so the only possible accumulation point is $0$ , and claims 1 and 2 are [step 3.1] and [step 2.2].

step 3.1step 2.2A4∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-09-22Open item page →

Fredholm operator cokernel and index

Definition

Let $X$ and $Y$ be Banach spaces over the same scalar field $F$ (Banach space) and let $T : X \to Y$ be a bounded linear operator (A bounded linear operator between normed spaces). Then $T$ is a Fredholm operator when all three of the following hold:

$\ker T$ is finite dimensional, that is, admits an ordered basis of finite length (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis);
$ran T$ is a closed subspace of $Y$ (Linear subspace of a vector space);
the cokernel $coker T : = Y / ran T$ is finite dimensional (The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M), Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

Here the cokernel is the algebraic quotient vector space; it also carries the quotient seminorm of The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M)), which is a norm when the range is closed (The quotient seminorm is a norm exactly when the subspace is closed), but the dimension in clause 3 is the vector-space dimension of the quotient and does not depend on that norm.

For a Fredholm operator $T$ the index of $T$ is the integer

$ind T : = \dim_{F} \ker T - \dim_{F} coker T$

Here subtraction is in $Z$ , not in $N$ : write $a = \dim_{F} \ker T$ and $b = \dim_{F} coker T$ and identify a natural $n$ with $[(n, 0)]$ . Precisely,

$ind T = [(a, b)] = [(a, 0)] + (- [(b, 0)]) \in Z$

by The integers as equivalence classes of pairs of naturals and Arithmetic on the integers. Both dimensions are natural numbers, so this class is defined even when $a < b$ ; the index may be positive, negative or zero.

Two remarks on the definition. The closedness of the range is listed explicitly as a hypothesis of the definition rather than extracted from the other clauses. And no property of the index is asserted here; additivity, local constancy and invariance under compact perturbations are separate theorems.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Fredholm splitting and parametrix

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ and $Y$ be Banach spaces over the same scalar field and let $T : X \to Y$ be a Fredholm operator (Fredholm operator cokernel and index, A bounded linear operator between normed spaces). Then there are a closed linear subspace $X_{1} \leq X$ and a finite-dimensional closed linear subspace $Y_{0} \leq Y$ with

$X = \ker T \oplus X_{1}, Y = ran T \oplus Y_{0},$

the coordinate projections of both decompositions being bounded (A complemented closed subspace of a normed space), with $\dim Y_{0} = \dim coker T$ (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis), and such that with $U : = (T ∣_{X_{1}})^{- 1} : ran T \to X_{1}$ , which is bounded, the operator

$S : Y ⟶ X, S (y) : = U (y_{1}) for y = y_{1} + y_{0}, y_{1} \in ran T, y_{0} \in Y_{0},$

is bounded and satisfies: $S T - I_{X}$ has finite-dimensional range of dimension at most $\dim \ker T$ , and $T S - I_{Y}$ has finite-dimensional range of dimension at most $\dim Y_{0}$ .

Facts & Assumptions

[A1]

A finite-dimensional linear subspace of a normed space is complemented, and a closed finite-codimensional linear subspace is complemented; a complemented subspace has a closed complement with bounded coordinate projections (Finite-dimensional subspaces are complemented, Closed finite-codimensional subspaces are complemented, A complemented closed subspace of a normed space, Linear subspace of a vector space).

[A2]

A closed linear subspace of a Banach space is a Banach space (A closed subspace of a Banach space is Banach, Banach space), and by the bounded inverse theorem, under DC, a bounded bijection between Banach spaces has a bounded inverse (Bounded inverse theorem); $A C$ supplies DC (AC supplies the countable and dependent choices used in Banach integration, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[A3]

If $q : Z \to Z / W$ is the quotient map and a linear bijection $Z_{0} \to Z / W$ is given, then choosing preimages of a finite basis is a finite selection (Every natural-number-indexed list of nonempty sets has a choice function on its family of values, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M), Linear map between vector spaces over the same field): a linearly independent spanning list pulls back to a linearly independent spanning list, because a linear bijection preserves the vanishing of finite linear combinations in both directions.

Proof

technique · direct

Given: $A C$ , Banach spaces $X, Y$ over one scalar field, a Fredholm operator $T : X \to Y$ with $N : = \ker T$ finite dimensional and $ran T$ closed with finite-dimensional cokernel.

1.1

There is a closed subspace $X_{1} \leq X$ with $X = N \oplus X_{1}$ and bounded projections.

1.2

There is a closed subspace $Y_{0} \leq Y$ with $Y = ran T \oplus Y_{0}$ and bounded projections; the quotient map $q : Y \to coker T = Y / ran T$ restricts to a linear bijection $q ∣_{Y_{0}} : Y_{0} \to coker T$ , which is injective because $Y_{0} \cap ran T = {0}$ and surjective because $y = y_{1} + y_{0}$ gives $q (y) = q (y_{0})$ .

2.1

The restriction $T_{1} : = T ∣_{X_{1}} : X_{1} \to ran T$ is a bounded linear bijection: it is injective because $X_{1} \cap \ker T = {0}$ , and surjective because $T (X) = T (N + X_{1}) = T (X_{1})$ .

step 1.1

2.2

The subspace $Y_{0}$ is finite dimensional with $\dim Y_{0} = \dim coker T$ : pulling back an ordered basis of the finite-dimensional quotient $coker T$ along the bijection $q ∣_{Y_{0}}$ of [step 1.2] gives an ordered basis of $Y_{0}$ , by the finite selection and independence argument of [A3].

step 1.2A3

3.1

The spaces $X_{1}$ and $ran T$ are Banach, so $U : = T_{1}^{- 1}$ is bounded by [A2].

step 2.1A2

4.1

The operator $S : Y \to X$ that equals $U$ on $ran T$ and $0$ on $Y_{0}$ is $U \circ P$ for the bounded projection $P : Y \to ran T$ of [step 1.2], hence bounded as a composite of bounded operators.

step 1.2step 3.1

4.2

For $x = n + x_{1}$ with $n \in N$ , $x_{1} \in X_{1}$ one has $S T x = S (T x_{1}) = U (T_{1} x_{1}) = x_{1}$ , so $S T$ is the bounded projection $P_{X_{1}}$ onto $X_{1}$ along $N$ and $S T - I_{X} = - P_{N}$ has range $N$ , of dimension $\dim \ker T$ .

step 1.1step 3.1

4.3

For $y = y_{1} + y_{0}$ one has $T S y = T (U y_{1}) = y_{1}$ , so $T S$ is the bounded projection $P$ onto $ran T$ along $Y_{0}$ and $T S - I_{Y}$ has range $Y_{0}$ , of dimension $\dim Y_{0}$ .

step 1.2step 3.1step 2.2

5.1

The decompositions, the boundedness of $U$ and $S$ and the two finite-rank defects are exactly the assertions, with $\dim Y_{0} = \dim coker T$ from [step 2.2].

step 4.1step 4.2step 2.2step 4.3∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

A compact remainder estimate forces closed range

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Let $X$ , $Y$ and $Z$ be Banach spaces over the same scalar field, let $T : X \to Y$ and $K : X \to Z$ be bounded linear operators with $K$ compact (A bounded linear operator between normed spaces, Compact linear operator), and suppose there is a real $C > 0$ with

$∥ x ∥ \leq C ∥ T x ∥ + ∥ K x ∥ for every x \in X .$

Then $\ker T$ is finite dimensional and $ran T$ is closed in $Y$ .

Facts & Assumptions

[A1]

Bounded linear operators are continuous, and the kernel of $T$ is a closed subspace of $X$ (For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent, A bounded linear operator between normed spaces); limits of sequences in a metric space are unique (A sequence in a metric space has at most one limit, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

[A2]

Assume DC. Then Countable Choice holds (Dependent choice implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )); a compact operator maps bounded sequences to sequences with convergent subsequences (Sequential characterization of compact operators, Sequences of reals: bounded, eventually, frequently, tails, subsequences, A strictly increasing index map satisfies $n_{k} \geq k$ ); a normed space with compact closed unit ball admits an ordered basis of finite length (The closed unit ball is compact if and only if the normed space is finite-dimensional); compact, sequentially compact and complete-and-totally-bounded agree for metric spaces under ${A C}_{ω}$ and DC (For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice).

[A3]

Under DC, $ran T$ is closed exactly when there is a real $C^{'} > 0$ with $dist (x, \ker T) \leq C^{'} ∥ T x ∥$ for every $x$ (Closed range is equivalent to a quotient estimate, The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))); the distance scales, $dist (λ x, \ker T) = ∣ λ ∣ dist (x, \ker T)$ , because $\ker T$ is a subspace, and $dist (u, M) \leq ∥ u ∥$ for nonempty $M ∋ 0$ .

[A4]

If $(u_{j})$ is Cauchy and some subsequence converges to $u$ , then $u_{j} \to u$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ); a closed set contains the limits of its convergent sequences (Banach space, A sequence in a metric space has at most one limit).

Proof

technique · direct

Given: $D C$ , Banach spaces $X, Y, Z$ over one scalar field, bounded $T : X \to Y$ , compact $K : X \to Z$ , a real $C > 0$ with $∥ x ∥ \leq C ∥ T x ∥ + ∥ K x ∥$ for all $x$ , and $N : = \ker T$ .

1.1

For $x \in N$ the estimate reads $∥ x ∥ \leq ∥ K x ∥$ , so $∥ K x - K x^{'} ∥ = ∥ K (x - x^{'}) ∥ \geq ∥ x - x^{'} ∥$ for all $x, x^{'} \in N$ .

algebra

1.2

The set $N$ is a closed subspace, hence a Banach space.

A1A4

1.3

For every real $η > 0$ there is $x$ with $dist (x, N) = 1$ , $∥ x ∥ \leq 2$ and $∥ T x ∥ < η$ whenever the estimate of [A3] fails for every constant: failure for the constant $C^{'} = 1 / η$ gives $x_{0}$ with $dist (x_{0}, N) > ∥ T x_{0} ∥ / η \geq 0$ and $dist (x_{0}, N) > 0$ , and choosing $m \in N$ with $∥ x_{0} - m ∥ < 2 dist (x_{0}, N)$ and setting $x : = (x_{0} - m) / dist (x_{0}, N)$ gives the three properties by scaling.

A3algebra

2.1

The closed unit ball $B_{N} : = N \cap {x : ∥ x ∥ \leq 1}$ is compact: if $(x_{j})$ is a sequence in $B_{N}$ , then it is bounded so by [A2] some subsequence has $K x_{j_{k}} \to z$ ; by [step 1.1] the subsequence is Cauchy, $∥ x_{j_{k}} - x_{j_{l}} ∥ \leq ∥ K x_{j_{k}} - K x_{j_{l}} ∥$ , hence converges to some $x \in N$ by [step 1.2], and $∥ x ∥ \leq 1$ ; thus every sequence in $B_{N}$ has a subsequence converging in $B_{N}$ , so $B_{N}$ is sequentially compact, hence compact by [A2].

step 1.1step 1.2A2

2.2

If the estimate of [A3] fails for every constant, then [step 1.3] makes the set of witnesses with $dist (w, N) = 1$ , $∥ w ∥ \leq 2$ and $∥ T w ∥ < 1 / (j + 1)$ nonempty for each $j$ ; Countable Choice in [A2] therefore supplies a sequence $(w_{j})$ with those three properties.

step 1.3A2

3.1

$\ker T$ is finite dimensional: its closed unit ball is compact by [step 2.1], so $N$ admits an ordered basis of finite length by [A2].

step 2.1A2

3.2

Under the hypothesis of [step 2.2] the bounded sequence $(w_{j})$ has, by [A2], a subsequence with $K w_{j_{k}} \to z$ for some $z \in Z$ .

step 2.2A2

4.1

Under the hypothesis of [step 2.2], the subsequence is Cauchy: $∥ w_{j_{k}} - w_{j_{l}} ∥ \leq C ∥ T (w_{j_{k}} - w_{j_{l}}) ∥ + ∥ K (w_{j_{k}} - w_{j_{l}}) ∥ \leq C (1 / (j_{k} + 1) + 1 / (j_{l} + 1)) + ∥ K w_{j_{k}} - K w_{j_{l}} ∥$ , and both terms tend to $0$ ; hence $w_{j_{k}} \to w$ for some $w \in X$ .

step 2.2step 3.2A4algebra

5.1

Under the hypothesis of [step 2.2], the limit $w$ lies in $N$ : $T w = \lim_{k} T w_{j_{k}} = 0$ by the continuity of $T$ and $∥ T w_{j_{k}} ∥ < 1 / (j_{k} + 1)$ .

step 2.2step 4.1A1A4

6.1

Under the hypothesis of [step 2.2], the numbers $dist (w_{j_{k}}, N) = 1$ converge to $dist (w, N)$ because the distance to a fixed set is 1-Lipschitz, so $dist (w, N) = 1$ , contradicting $w \in N$ of [step 5.1], which forces $dist (w, N) = 0$ .

step 2.2step 4.1step 5.1A3

7.1

Hence the estimate of [A3] holds for some constant, and then $ran T$ is closed by [A3]; together with [step 3.1] this proves the lemma.

step 3.1step 6.1A3∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Atkinson

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ and $Y$ be Banach spaces over the same scalar field and let $T : X \to Y$ be a bounded linear operator (A bounded linear operator between normed spaces). Then $T$ is Fredholm (Fredholm operator cokernel and index) if and only if there is a bounded linear $S : Y \to X$ such that both $S T - I_{X}$ and $T S - I_{Y}$ are compact (Compact linear operator).

Facts & Assumptions

[A1]

If $T$ is Fredholm, the splitting lemma provides a bounded $S : Y \to X$ for which $S T - I_{X}$ has finite-dimensional range of dimension at most $\dim \ker T$ and $T S - I_{Y}$ has finite-dimensional range of dimension at most $\dim coker T$ (Fredholm splitting and parametrix); a bounded finite-rank operator is compact (Bounded finite rank operators are compact, Fredholm operator cokernel and index).

[A2]

Under DC, if $∥ x ∥ \leq C ∥ T x ∥ + ∥ K x ∥$ for a compact $K$ and some real $C > 0$ , then $\ker T$ is finite dimensional and $ran T$ is closed (A compact remainder estimate forces closed range, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, Banach space); $A C$ supplies DC (AC supplies the countable and dependent choices used in Banach integration).

[A3]

Transposition is additive with $(B A)^{*} = A^{*} B^{*}$ and $I^{*} = I$ (Transposition reverses composition, The transpose of a bounded operator); a compact operator between Banach spaces has compact transpose (Schauder compact adjoint theorem); the kernel of $I - C$ with $C$ compact is finite dimensional (Kernel of identity minus compact is finite dimensional).

[A4]

For a bounded $T$ , $(ran T)^{⊥} = \ker T^{*}$ and $\overline{ran T} =^{⊥} (\ker T^{*})$ (Elementary kernel and range annihilator identities); for closed $M \leq Y$ the map $(Y / M)^{*} \to M^{⊥}$ , $h \mapsto h \circ q$ , is a linear isometric bijection (The dual of a quotient is its annihilator, The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M)).

[A5]

For nonzero $z$ in a normed space $Z$ there is $h \in Z^{*}$ with $∥ h ∥ = 1$ and $h (z) = ∥ z ∥$ (Every nonzero vector has a norming functional), so the dual separates points (The dual space separates points of a normed space); a subspace of a finite-dimensional space is finite dimensional (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ , Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis), and a linear bijection carries an ordered basis to an ordered basis (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

Proof

technique · direct

Given: $A C$ , Banach spaces $X, Y$ over one scalar field, and a bounded linear $T : X \to Y$ .

1.1

If $T$ is Fredholm, then the operator $S$ of [A1] satisfies: $S T - I_{X}$ and $T S - I_{Y}$ have finite-dimensional ranges, hence are compact.

1.2

Conversely assume there is a bounded $S : Y \to X$ with $F : = S T - I_{X}$ and $G : = T S - I_{Y}$ compact. By boundedness choose a real $b \geq 0$ such that $∥ S y ∥ \leq b ∥ y ∥$ for every $y \in Y$ , and put $C : = \max (b, 1) > 0$ .

assume-hyp

2.1

For every $x \in X$ one has $x = S T x - F x$ , hence $∥ x ∥ \leq ∥ S T x ∥ + ∥ F x ∥ \leq b ∥ T x ∥ + ∥ F x ∥ \leq C ∥ T x ∥ + ∥ F x ∥$ .

step 1.2algebra

2.2

By [A3] the transpose of $G = T S - I_{Y}$ is $G^{*} = S^{*} T^{*} - I_{Y^{*}}$ , so $S^{*} T^{*} = I_{Y^{*}} + G^{*} = I_{Y^{*}} - (- G^{*})$ ; for $g \in Y^{*}$ with $T^{*} g = 0$ one has $S^{*} T^{*} g = S^{*} (T^{*} g) = 0$ by linearity of $S^{*}$ , hence $(I_{Y^{*}} - (- G^{*})) g = S^{*} T^{*} g = 0$ and $\ker T^{*} \subseteq \ker (I_{Y^{*}} - (- G^{*}))$ .

step 1.2A3

3.1

Under the hypothesis of [step 1.2], $\ker T$ is finite dimensional and $ran T$ is closed, by [A2] applied to the estimate of [step 2.1] with the compact operator $F$ .

step 2.1A2

3.2

Under the hypothesis of [step 1.2] the operator $G : = T S - I_{Y}$ is compact, so its transpose $G^{*}$ is compact by [A3]; the negative $- G^{*}$ is compact as well, because the image of a bounded set under $- G^{*}$ is the negative of its image under $G^{*}$ and negating a set preserves the compactness of its closure. So [A3] applies to the compact operator $- G^{*}$ and makes $\ker (I_{Y^{*}} - (- G^{*}))$ finite dimensional; by [step 2.2] the subspace $\ker T^{*}$ is finite dimensional.

step 1.2step 2.2A3

4.1

Under the hypothesis of [step 1.2], the dual of the cokernel is finite dimensional: since $ran T$ is closed by [step 3.1], [A4] gives $(coker T)^{*} = (Y / ran T)^{*} ≅ (ran T)^{⊥} = \ker T^{*}$ , which is finite dimensional by [step 3.2].

step 3.1step 3.2A4

5.1

Under the hypothesis of [step 1.2], the cokernel is finite dimensional: if $Z$ is a normed space whose dual has ordered basis $h_{1}, \dots, h_{n}$ , then $Ψ (z) : = (h_{1} (z), \dots, h_{n} (z))$ is linear and injective, because a nonzero $z$ has by [A5] a norm-one functional $h$ , and $h = \sum_{i} c_{i} h_{i}$ forces $Ψ (z) \neq 0$ ; the inverse bijection carries an ordered basis of the finite-dimensional image $Ψ (Z)$ to an ordered basis of $Z$ by [A5].

step 4.1A5

6.1

Under the hypothesis of [step 1.2] the operator $T$ is Fredholm, since its kernel is finite dimensional by [step 3.1], its range is closed by [step 3.1] and its cokernel is finite dimensional by [step 5.1]; with [step 1.1] this is the asserted equivalence.

step 1.1step 3.1step 5.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Fredholm index is additive

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ , $Y$ and $Z$ be Banach spaces over the same scalar field, and let $T : X \to Y$ and $U : Y \to Z$ be Fredholm operators (Fredholm operator cokernel and index, A bounded linear operator between normed spaces). Then $U T : X \to Z$ is Fredholm and

$ind (U T) = ind U + ind T .$

Facts & Assumptions

[A1]

By Atkinson's theorem a bounded $A$ is Fredholm exactly when there is a bounded $B$ with $A B - I$ and $B A - I$ compact (Atkinson); the compact operators are closed under sums, scalar multiples and composition with bounded operators (Linear combinations of compact operators are compact, Compositions with a compact operator are compact, Fredholm operator cokernel and index).

[A2]

Rank-nullity: for a linear map $f : V \to W$ with $V$ finite dimensional, $\dim V = \dim \ker f + \dim ran f$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ , Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis); and $\dim \ker f = \dim im g$ for an exact predecessor $g$ at that spot, so that a finite exact sequence $0 \to V_{1} \to V_{2} \to V_{3} \to V_{4} \to V_{5} \to V_{6} \to 0$ of finite-dimensional spaces satisfies $\sum_{i = 1}^{6} (- 1)^{i + 1} \dim V_{i} = 0$ (Linear map between vector spaces over the same field, Linear subspace of a vector space).

[A3]

For a bounded $T$ the cokernel is the quotient $coker T = Y / ran T$ with cosets written $y + ran T$ (The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M), Fredholm operator cokernel and index); $A C$ supplies DC (AC supplies the countable and dependent choices used in Banach integration, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, Banach space).

Proof

technique · direct

Given: $A C$ , Banach spaces $X, Y, Z$ over one scalar field, Fredholm operators $T : X \to Y$ and $U : Y \to Z$ , and parametrices $S$ for $T$ and $R$ for $U$ as in [A1].

1.1

$U T$ is Fredholm: $S R$ is a parametrix for $U T$ , since $(S R) (U T) - I_{X} = S (R U - I_{Y}) T + (S T - I_{X})$ and $(U T) (S R) - I_{Z} = U (T S - I_{Y}) R + (U R - I_{Z})$ are compact by [A1], so Atkinson gives Fredholmness of $U T$ .

1.2

The maps $α : V_{1} \to V_{2}$ , $α (x) = x$ ; $β : V_{2} \to V_{3}$ , $β (x) = T x$ ; $γ : V_{3} \to V_{4}$ , $γ (y) = y + ran T$ ; $δ : V_{4} \to V_{5}$ , $δ (y + ran T) = U y + ran (U T)$ ; and $ε : V_{5} \to V_{6}$ , $ε (z + ran (U T)) = z + ran U$ are well-defined linear maps: $δ$ is well-defined because $y \in ran T$ gives $U y \in ran (U T)$ , and the other four are restrictions, inclusions or quotient maps of linear maps.

2.1

All six spaces $V_{1} : = \ker T$ , $V_{2} : = \ker (U T)$ , $V_{3} : = \ker U$ , $V_{4} : = coker T$ , $V_{5} : = coker (U T)$ , $V_{6} : = coker U$ are finite dimensional.

step 1.1A1

2.2

The sequence is exact at $V_{1}$ , $V_{2}$ and $V_{3}$ : $α$ is injective; $\ker β = {x \in V_{2} : T x = 0} = V_{1} = im α$ ; and $\ker γ = {y \in V_{3} : y \in ran T} = {T x : x \in V_{2}} = im β$ .

step 1.2

2.3

The sequence is exact at $V_{4}$ , $V_{5}$ and $V_{6}$ : $\ker δ = {y + ran T : U y \in ran (U T)} = {y + ran T : y - T x \in \ker U for some x} = im γ$ ; $im δ = {U y + ran (U T) : y \in Y} = {z + ran (U T) : z \in ran U} = \ker ε$ ; and $ε$ is surjective as the quotient map $Z / ran (U T) \to Z / ran U$ .

step 1.2

3.1

Rank-nullity telescopes the dimensions: with $f_{0} : 0 \to V_{1}$ , $f_{i} : V_{i} \to V_{i + 1}$ for $1 \leq i \leq 5$ and $f_{6} : V_{6} \to 0$ the maps of [step 1.2], exactness gives $\ker f_{i} = im f_{i - 1}$ , so $\dim V_{i} = \dim im f_{i - 1} + \dim im f_{i}$ , and summing with signs $(+, -, +, -, +, -)$ cancels to $\dim V_{1} - \dim V_{2} + \dim V_{3} - \dim V_{4} + \dim V_{5} - \dim V_{6} = 0$ , because $im f_{0} = 0$ and $im f_{6} = 0$ .

step 2.1step 2.2step 2.3A2

4.1

The index identity follows: $ind (U T) = \dim V_{2} - \dim V_{5} = (\dim V_{1} - \dim V_{4}) + (\dim V_{3} - \dim V_{6}) = ind T + ind U$ , by the telescoping identity of [step 3.1] and the definition of the index.

step 3.1A2A3∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Fredholm index is locally constant

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ and $Y$ be Banach spaces over the same scalar field. Then the Fredholm operators $X \to Y$ (Fredholm operator cokernel and index) form an open subset of the space $B (X, Y)$ of bounded linear operators with the operator norm (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum): for every Fredholm $T$ there is a real $δ > 0$ such that every bounded $A : X \to Y$ with $∥ A - T ∥ < δ$ is Fredholm, and then $ind A = ind T$ .

Facts & Assumptions

[A1]

A Fredholm $T$ admits bounded projections splitting $X = \ker T \oplus X_{1}$ and $Y = ran T \oplus Y_{0}$ , with $N : = \ker T$ and $Y_{0}$ finite dimensional, $\dim Y_{0} = \dim coker T$ , and with $T_{1} : = T ∣_{X_{1}} : X_{1} \to ran T$ a bounded isomorphism whose inverse $T_{1}^{- 1}$ is bounded (Fredholm splitting and parametrix).

[A2]

Neumann: if $∥ T_{1}^{- 1} ∥ ∥ C ∥ < 1$ then $T_{1} + C$ is invertible with bounded inverse (Neumann series and small perturbations of bounded inverses, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A3]

Fredholm operators are closed under composition between Banach spaces and the index is additive, $ind (U T) = ind U + ind T$ (Fredholm index is additive, Fredholm operator cokernel and index); an invertible bounded operator is Fredholm with index $0$ , its kernel and cokernel being ${0}$ .

[A4]

A linear map defined on a finite-dimensional normed space is bounded (A linear map from a finite-dimensional normed space is bounded); rank-nullity (Rank-nullity: $\dim_{F} V = nullity T + rank T$ , Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis); and for a block-diagonal operator $diag (A_{11}, S)$ on $ran T \oplus Y_{0}$ the kernel is $\ker A_{11} \oplus \ker S$ and the cokernel is isomorphic to $coker A_{11} \oplus coker S$ (The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M), Linear subspace of a vector space).

Proof

technique · direct

Given: $A C$ , Banach spaces $X, Y$ over one scalar field, a Fredholm $T : X \to Y$ , and the splitting $X = N \oplus X_{1}$ , $Y = ran T \oplus Y_{0}$ of [A1].

1.1

The projections $P_{N}$ , $P_{X_{1}}$ , $P_{ran T}$ , $P_{Y_{0}}$ of the two splittings are bounded; write $κ : = \max (1, ∥ P_{ran T} ∥, ∥ P_{Y_{0}} ∥)$ .

1.2

If $X_{1} = {0}$ , then $ran T = {0}$ , so $X = N$ and $Y = Y_{0}$ are finite dimensional, $T = 0$ , and for every bounded $A : X \to Y$ rank-nullity gives $ind A = \dim \ker A - \dim coker A = \dim X - \dim Y = ind T$ ; so the claim holds with any $δ > 0$ in this case.

A1A4

2.1

Assume $X_{1} \neq {0}$ , so $ran T \neq {0}$ and $∥ T_{1}^{- 1} ∥ > 0$ , and put $δ : = 1 / (2 κ ∥ T_{1}^{- 1} ∥) > 0$ . For every bounded $A$ with $∥ A - T ∥ < δ$ , writing $A_{11} = P_{ran T} A ∣_{X_{1}}$ and $T_{11} = T_{1}$ one has $∥ A_{11} - T_{1} ∥ \leq κ ∥ A - T ∥$ , so $∥ T_{1}^{- 1} (A_{11} - T_{1}) ∥ < 1 / 2 < 1$ and $A_{11} = T_{1} (I + T_{1}^{- 1} (A_{11} - T_{1}))$ is invertible with bounded inverse by [A2].

step 1.1A1A2algebra

3.1

Under the hypothesis of [step 2.1], reorder the domain splitting as $X = X_{1} \oplus N$ and keep the codomain splitting $Y = ran T \oplus Y_{0}$ . Let $U : Y \to Y$ and $V : X \to X$ be the bounded operators whose block matrices in these stated orders are $U = (\begin{array}{cc} I_{ran T} & 0 \\ - A_{21} A_{11}^{- 1} & I_{Y_{0}} \end{array})$ and $V = (\begin{array}{cc} I_{X_{1}} & - A_{11}^{- 1} A_{12} \\ 0 & I_{N} \end{array})$ , where $A_{12} = P_{ran T} A ∣_{N}$ , $A_{21} = P_{Y_{0}} A ∣_{X_{1}}$ and $A_{22} = P_{Y_{0}} A ∣_{N}$ ; then $U A V = diag (A_{11}, S)$ with $S : = A_{22} - A_{21} A_{11}^{- 1} A_{12}$ , and $U, V$ are invertible with bounded inverses given by the same matrices with the off-diagonal signs reversed.

step 1.1step 2.1A1algebra

4.1

Under the hypothesis of [step 2.1], $U A V = diag (A_{11}, S)$ is Fredholm with index $ind (U A V) = ind A_{11} + ind S = ind S$ , because $A_{11}$ is an isomorphism of $X_{1}$ onto $ran T$ and $S$ maps the finite-dimensional space $N$ boundedly into the finite-dimensional space $Y_{0}$ ; by [A4] its index is $ind S = \dim \ker S - \dim coker S = \dim \ker S + \dim im S - \dim Y_{0} = \dim N - \dim Y_{0}$ .

step 3.1A4

5.1

Under the hypothesis of [step 2.1], $A$ is Fredholm with $ind A = \dim N - \dim Y_{0} = ind T$ : since $U, V$ and their inverses are invertible hence Fredholm of index $0$ , [A3] gives first that $A = U^{- 1} (U A V) V^{- 1}$ is Fredholm, and then $ind A = ind (U^{- 1}) + ind (U A V) + ind (V^{- 1}) = ind (U A V)$ .

step 3.1step 4.1A3

6.1

In the case of [step 1.2] and in the case of [step 5.1] every bounded $A$ with $∥ A - T ∥$ below the corresponding $δ$ (any positive number in the first case, the $δ$ of [step 2.1] in the second) is Fredholm of index $ind T$ , so the Fredholm operators are open in $B (X, Y)$ and the index is locally constant at $T$ .

step 1.2step 5.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Fredholm index is stable under compact perturbations

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ and $Y$ be Banach spaces over the same scalar field, let $T : X \to Y$ be a Fredholm operator and let $K : X \to Y$ be compact (Fredholm operator cokernel and index, Compact linear operator, A bounded linear operator between normed spaces). Then $T + K$ is Fredholm and $ind (T + K) = ind T$ .

Facts & Assumptions

[A1]

By Atkinson's theorem a bounded operator is Fredholm exactly when it has a bounded parametrix modulo compact operators (Atkinson); compact operators are closed under sums, scalar multiples and composition with bounded operators (Linear combinations of compact operators are compact, Compositions with a compact operator are compact, Fredholm operator cokernel and index).

[A2]

The Fredholm operators form an open subset of $B (X, Y)$ and the index is locally constant (Fredholm index is locally constant, The operator norm as the least bound and as the unit-sphere or unit-ball supremum): for each Fredholm $A$ there is $δ_{A} > 0$ such that every bounded $B$ with $∥ B - A ∥ < δ_{A}$ is Fredholm with $ind B = ind A$ .

[A3]

The interval $[0, 1]$ is a connected subset of $R$ (A subset of $R$ is connected if and only if it is order-convex, that is, an interval, Separated sets, disconnection, and connected subset of $R$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). A real point lies in the closure of a set exactly when each of its neighbourhoods meets that set (The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points). Convergence in operator norm is metric convergence (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , Open ball, closed ball and sphere in a metric space).

Proof

technique · direct

Given: $A C$ , Banach spaces $X, Y$ over one scalar field, a Fredholm $T : X \to Y$ , a compact $K : X \to Y$ , and a parametrix $S$ for $T$ with $S T - I_{X}$ and $T S - I_{Y}$ compact.

1.1

For every $t \in [0, 1]$ the operator $T + t K$ is Fredholm: $S$ is a parametrix for it modulo compact operators, because $S (T + t K) - I_{X} = (S T - I_{X}) + t S K$ and $(T + t K) S - I_{Y} = (T S - I_{Y}) + t K S$ are compact by [A1], so Atkinson applies.

1.2

The path $t \mapsto T + t K$ is continuous for the operator norm: $∥ (T + s K) - (T + t K) ∥ \leq ∣ s - t ∣ ∥ K ∥$ for all real $s, t$ .

A1A3algebra

2.1

Put $U : = {t \in [0, 1] : ind (T + t K) = ind T}$ . For every $t \in U$ , local constancy [A2] gives $δ > 0$ with all operators within $δ$ of $T + t K$ having the same index. With $η : = δ / (1 + ∥ K ∥) > 0$ , every $s \in [0, 1]$ satisfying $∣ s - t ∣ < η$ has $∥ (T + s K) - (T + t K) ∥ \leq ∣ s - t ∣ ∥ K ∥ < δ$ , hence lies in $U$ .

step 1.2A2

2.2

Put $V : = [0, 1] ∖ U$ . For every $t \in V$ , the same argument gives $η > 0$ such that every $s \in [0, 1]$ with $∣ s - t ∣ < η$ has index $ind (T + t K) \neq ind T$ and hence lies in $V$ .

step 1.2A2

3.1

Hence $U = [0, 1]$ . Indeed $0 \in U$ . If $V$ were nonempty, then $U \cup V = [0, 1]$ would be a disconnection: if $t \in V$ , step 2.2 supplies a neighbourhood of $t$ whose intersection with $[0, 1]$ is contained in $V$ , so this neighbourhood misses $U$ and [A3] gives $t \notin \overline{U}$ ; hence $\overline{U} \cap V = \emptyset$ . Similarly step 2.1 gives $U \cap \overline{V} = \emptyset$ . Thus the two nonempty sets would be separated, contradicting connectedness of $[0, 1]$ in [A3].

step 2.1step 2.2A3

4.1

In particular $1 \in U$ , so $T + K$ is Fredholm with $ind (T + K) = ind T$ , as claimed.

step 1.1step 3.1∎

CorollaryStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Lambda identity minus compact has index zero

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ be a Banach space over $R$ or $C$ , let $K : X \to X$ be a compact operator (Compact linear operator) and let $λ \neq 0$ be a scalar. Then $λ I - K$ is a Fredholm operator and $ind (λ I - K) = 0$ (Fredholm operator cokernel and index).

Facts & Assumptions

[A1]

If $A$ is Fredholm and $C$ is compact then $A + C$ is Fredholm with $ind (A + C) = ind A$ (Fredholm index is stable under compact perturbations).

[A2]

A scalar multiple of a compact operator is compact (Linear combinations of compact operators are compact); the identity is bounded linear, $λ I$ is invertible with inverse $λ^{- 1} I$ for $λ \neq 0$ , and an invertible bounded operator is Fredholm of index $0$ , because its kernel and cokernel are the zero spaces (Linear map between vector spaces over the same field, A bounded linear operator between normed spaces, Fredholm operator cokernel and index, The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M), Banach space).

Proof

technique · direct

Given: $A C$ , a Banach space $X$ over $R$ or $C$ , a compact $K : X \to X$ and a scalar $λ \neq 0$ .

1.1

The operator $λ I$ is bounded and invertible with inverse $λ^{- 1} I$ , hence Fredholm with $ind (λ I) = \dim {0} - \dim {0} = 0$ .

1.2

The operator $- K$ is compact by [A2], and $λ I - K = λ I + (- K)$ is a compact perturbation of the Fredholm operator $λ I$ .

A2algebra

2.1

By [A1] the operator $λ I - K$ is Fredholm and $ind (λ I - K) = ind (λ I) = 0$ , which is the claim.

step 1.1step 1.2A1∎

5 · Examples, counterexamples and false statements

None yet.